Stoichiometry
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- Exercise 1
- Exercise 2
- Chemical Equations and Stoichiometry
- Exercise 3
- Exercise 4
- Exercise 5
- Exercise 6 - Part a
- Exercise 6 - Part b
- Exercise 6 - Part c
- Limiting Reactants
- Stoichiometry in Solutions
- Exercise 7
- Exercise 8
- Exercise 9
- Exercise 10
- Exercise 11
- Exercise 12
- Diluting Solutions
- Exercise 13
- Exercise 14
- Exercise 15
- Exercise 16
- Exercise 17
- Exercise 18
- Exercise 19
- Exercise 20
- Exercise 21
- Reaction Yields - Theory
- Reaction Yields - Example
- Exercise 22

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[{"Name":"Stoichiometry","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 1","Duration":"9m 21s","ChapterTopicVideoID":22857,"CourseChapterTopicPlaylistID":86817,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22857.jpeg","UploadDate":"2020-12-15T05:56:30.5470000","DurationForVideoObject":"PT9M21S","Description":null,"VideoComments":[],"Subtitles":[],"ID":23678},{"Watched":false,"Name":"Exercise 2","Duration":"7m 25s","ChapterTopicVideoID":22859,"CourseChapterTopicPlaylistID":86817,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22859.jpeg","UploadDate":"2020-12-15T05:57:25.1970000","DurationForVideoObject":"PT7M25S","Description":null,"VideoComments":[],"Subtitles":[],"ID":23680},{"Watched":false,"Name":"Chemical Equations and Stoichiometry","Duration":"11m 49s","ChapterTopicVideoID":16923,"CourseChapterTopicPlaylistID":86817,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/16923.jpeg","UploadDate":"2019-02-20T23:49:42.4470000","DurationForVideoObject":"PT11M49S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.520","Text":"In a previous video,"},{"Start":"00:02.520 ","End":"00:07.050","Text":"we wrote the chemical equation for the combustion of glucose."},{"Start":"00:07.050 ","End":"00:09.600","Text":"In this video, we\u0027ll see that we can use"},{"Start":"00:09.600 ","End":"00:13.830","Text":"the balanced equation to solve chemical problems."},{"Start":"00:13.830 ","End":"00:17.640","Text":"What does this chemical equation tell us?"},{"Start":"00:17.640 ","End":"00:20.565","Text":"When we burn 1 mole of glucose,"},{"Start":"00:20.565 ","End":"00:21.840","Text":"this should be a 1 here,"},{"Start":"00:21.840 ","End":"00:23.130","Text":"but we don\u0027t write it,"},{"Start":"00:23.130 ","End":"00:25.290","Text":"and 6 moles of oxygen,"},{"Start":"00:25.290 ","End":"00:26.640","Text":"here\u0027s the 6,"},{"Start":"00:26.640 ","End":"00:32.115","Text":"we get 6 moles of carbon dioxide and 6 moles of water."},{"Start":"00:32.115 ","End":"00:35.940","Text":"This is result of chemical bookkeeping."},{"Start":"00:35.940 ","End":"00:43.595","Text":"It tells us how many moles we use up in order to produce a certain number of moles."},{"Start":"00:43.595 ","End":"00:46.370","Text":"It\u0027s called stoichiometry."},{"Start":"00:46.370 ","End":"00:51.290","Text":"Stoichiometry, difficult word to spell."},{"Start":"00:51.290 ","End":"00:57.620","Text":"The chemical equation we had for glucose is a simple example of stoichiometry."},{"Start":"00:57.620 ","End":"01:02.299","Text":"Stoichiometry tells us how to perform"},{"Start":"01:02.299 ","End":"01:07.415","Text":"quantitative calculations based on chemical equations."},{"Start":"01:07.415 ","End":"01:09.230","Text":"In order to do this,"},{"Start":"01:09.230 ","End":"01:11.465","Text":"in order to solve problems,"},{"Start":"01:11.465 ","End":"01:14.975","Text":"we need to write conversion factors based,"},{"Start":"01:14.975 ","End":"01:18.020","Text":"for example, on the combustion of glucose."},{"Start":"01:18.020 ","End":"01:21.040","Text":"Let\u0027s write conversion factors."},{"Start":"01:21.040 ","End":"01:23.100","Text":"From 1 mole of glucose,"},{"Start":"01:23.100 ","End":"01:25.545","Text":"we get 6 moles of carbon dioxide."},{"Start":"01:25.545 ","End":"01:29.950","Text":"We can make that into a ratio and write this is equal to 1,"},{"Start":"01:29.950 ","End":"01:36.755","Text":"1 mole of glucose is equivalent in some respect to 6 moles of carbon dioxide."},{"Start":"01:36.755 ","End":"01:44.485","Text":"Similarly, 1 mole of glucose is equivalent to 6 moles of water."},{"Start":"01:44.485 ","End":"01:48.050","Text":"Burning 1 mole of glucose gives us 6 moles of water."},{"Start":"01:48.050 ","End":"01:52.130","Text":"We can make that into a ratio and write equal to 1."},{"Start":"01:52.130 ","End":"01:58.085","Text":"Similarly, 6 moles of water is equivalent to 6 moles of carbon dioxide."},{"Start":"01:58.085 ","End":"01:59.895","Text":"We can divide the 6,"},{"Start":"01:59.895 ","End":"02:06.250","Text":"and write 1 mole of water is equivalent to 1 mole of carbon dioxide."},{"Start":"02:06.250 ","End":"02:09.050","Text":"All these are conversion factors."},{"Start":"02:09.050 ","End":"02:13.190","Text":"We call them stoichiometric ratios."},{"Start":"02:13.190 ","End":"02:17.585","Text":"We can insert them wherever it is relevant."},{"Start":"02:17.585 ","End":"02:19.624","Text":"Let\u0027s take an example."},{"Start":"02:19.624 ","End":"02:23.390","Text":"How many moles of carbon dioxide are produced when"},{"Start":"02:23.390 ","End":"02:28.655","Text":"1.5 moles of glucose is burnt in excess oxygen."},{"Start":"02:28.655 ","End":"02:30.680","Text":"That means we\u0027ve got so much oxygen,"},{"Start":"02:30.680 ","End":"02:32.830","Text":"we don\u0027t need to worry about it."},{"Start":"02:32.830 ","End":"02:36.910","Text":"The number of moles of carbon dioxide is equal to the number of"},{"Start":"02:36.910 ","End":"02:42.895","Text":"moles of glucose multiplied by our conversion factor,"},{"Start":"02:42.895 ","End":"02:46.090","Text":"our stoichiometric ratio, 6 moles"},{"Start":"02:46.090 ","End":"02:49.630","Text":"of carbon dioxide being equivalent to 1 mole of glucose."},{"Start":"02:49.630 ","End":"02:52.240","Text":"First, let\u0027s multiply the numbers."},{"Start":"02:52.240 ","End":"02:55.930","Text":"We start off with 1.5 moles of glucose,"},{"Start":"02:55.930 ","End":"02:59.610","Text":"here it\u0027s 1.5 and here we have 6 divide by 1,"},{"Start":"02:59.610 ","End":"03:01.055","Text":"so we have 6."},{"Start":"03:01.055 ","End":"03:03.085","Text":"Now we need the units."},{"Start":"03:03.085 ","End":"03:06.670","Text":"We have moles of glucose divided by moles of glucose,"},{"Start":"03:06.670 ","End":"03:08.290","Text":"so we can cancel those,"},{"Start":"03:08.290 ","End":"03:11.690","Text":"and we\u0027re left with moles of carbon dioxide."},{"Start":"03:11.820 ","End":"03:16.060","Text":"Multiply 1.5 times 6 and we get,"},{"Start":"03:16.060 ","End":"03:17.660","Text":"of course 9,"},{"Start":"03:17.660 ","End":"03:20.190","Text":"9 moles of carbon dioxide."},{"Start":"03:20.190 ","End":"03:22.295","Text":"That\u0027s the solution to the problem."},{"Start":"03:22.295 ","End":"03:26.029","Text":"When we burn 1.5 moles of glucose,"},{"Start":"03:26.029 ","End":"03:29.300","Text":"we get 9 moles of carbon dioxide."},{"Start":"03:29.300 ","End":"03:31.880","Text":"That\u0027s a very simple problem."},{"Start":"03:31.880 ","End":"03:37.880","Text":"We can also write it and we\u0027ll use this notation in the rest of the video."},{"Start":"03:37.880 ","End":"03:40.460","Text":"Number of moles of carbon dioxide,"},{"Start":"03:40.460 ","End":"03:47.590","Text":"that\u0027s n for numbers of moles and carbon dioxide indicates of what is equal to 9 moles."},{"Start":"03:47.590 ","End":"03:53.150","Text":"We could also write 9 moles of carbon dioxide if we want it to be very exact."},{"Start":"03:53.150 ","End":"03:59.100","Text":"Often the calculations are more complex than just what we had before."},{"Start":"03:59.100 ","End":"04:05.570","Text":"But the central part always involves stoichiometric conversion factors."},{"Start":"04:05.570 ","End":"04:09.710","Text":"The central part is always converting moles to moles,"},{"Start":"04:09.710 ","End":"04:12.380","Text":"moles of products to moles of reactants,"},{"Start":"04:12.380 ","End":"04:14.705","Text":"moles of reactant to moles of products,"},{"Start":"04:14.705 ","End":"04:16.685","Text":"depending on the question."},{"Start":"04:16.685 ","End":"04:19.760","Text":"Let\u0027s write it out like this."},{"Start":"04:19.760 ","End":"04:22.280","Text":"The information we\u0027re given about the reactants,"},{"Start":"04:22.280 ","End":"04:24.965","Text":"we have to convert to the moles of reactants."},{"Start":"04:24.965 ","End":"04:28.865","Text":"Then the moles of reactants have to be converted to the moles of products."},{"Start":"04:28.865 ","End":"04:31.580","Text":"That\u0027s a central part of the problem."},{"Start":"04:31.580 ","End":"04:34.875","Text":"The moles of products have to be converted to"},{"Start":"04:34.875 ","End":"04:38.915","Text":"the information required about the products."},{"Start":"04:38.915 ","End":"04:40.535","Text":"Let\u0027s take an example."},{"Start":"04:40.535 ","End":"04:48.290","Text":"What volume of water is produced when 270 grams of glucose is burnt in excess oxygen."},{"Start":"04:48.290 ","End":"04:54.815","Text":"Let\u0027s convert the scheme that we had above into something a little bit more precise."},{"Start":"04:54.815 ","End":"04:57.935","Text":"We\u0027re given the mass of glucose."},{"Start":"04:57.935 ","End":"05:03.920","Text":"We want to convert it into the moles of glucose."},{"Start":"05:03.920 ","End":"05:06.890","Text":"This is always the central part of the problem."},{"Start":"05:06.890 ","End":"05:11.045","Text":"Then we have to convert it into the moles of water."},{"Start":"05:11.045 ","End":"05:12.920","Text":"From the moles of water,"},{"Start":"05:12.920 ","End":"05:16.295","Text":"we need to convert it into the mass of water,"},{"Start":"05:16.295 ","End":"05:20.255","Text":"and from the mass of water to the volume of water,"},{"Start":"05:20.255 ","End":"05:21.965","Text":"which is what we\u0027re asked about."},{"Start":"05:21.965 ","End":"05:24.440","Text":"We\u0027re asked what volume of water."},{"Start":"05:24.440 ","End":"05:27.785","Text":"We can label this 1,"},{"Start":"05:27.785 ","End":"05:31.080","Text":"2, 3, 4."},{"Start":"05:31.080 ","End":"05:33.235","Text":"These are the 4 steps."},{"Start":"05:33.235 ","End":"05:37.685","Text":"Before we can start to solve this problem."},{"Start":"05:37.685 ","End":"05:42.500","Text":"We need to recall some things that we learned before."},{"Start":"05:42.500 ","End":"05:47.930","Text":"The first thing we need to remember is how to convert the mass to number of moles."},{"Start":"05:47.930 ","End":"05:49.790","Text":"We need the following equation."},{"Start":"05:49.790 ","End":"05:55.790","Text":"We need the number of moles is equal to the mass divided by the molar mass."},{"Start":"05:55.790 ","End":"05:58.925","Text":"Mw, we\u0027re going to write as molar mass."},{"Start":"05:58.925 ","End":"06:02.355","Text":"I know it doesn\u0027t sound right. It\u0027s Mw."},{"Start":"06:02.355 ","End":"06:07.790","Text":"It\u0027s because once we used to call it molecular weight and the name has stuck."},{"Start":"06:07.790 ","End":"06:10.040","Text":"But it\u0027s n = m,"},{"Start":"06:10.040 ","End":"06:12.560","Text":"the mass divided by the molar mass,"},{"Start":"06:12.560 ","End":"06:15.035","Text":"so Mw is molar mass."},{"Start":"06:15.035 ","End":"06:19.520","Text":"At the end, we need to convert the number of moles to mass."},{"Start":"06:19.520 ","End":"06:22.130","Text":"That\u0027s the opposite of what we\u0027ve just done."},{"Start":"06:22.130 ","End":"06:23.930","Text":"We take this equation,"},{"Start":"06:23.930 ","End":"06:27.100","Text":"n = m divided by Mw."},{"Start":"06:27.100 ","End":"06:28.950","Text":"We want m,"},{"Start":"06:28.950 ","End":"06:37.175","Text":"so we can multiply both sides by Mw and we get the mass is equal to n times Mw."},{"Start":"06:37.175 ","End":"06:42.005","Text":"These are just 2 versions of the same equation."},{"Start":"06:42.005 ","End":"06:43.970","Text":"Now before we can proceed,"},{"Start":"06:43.970 ","End":"06:48.035","Text":"we need to know what the molar masses are and we can easily calculate"},{"Start":"06:48.035 ","End":"06:54.535","Text":"the molar mass of glucose is equal to 180 grams."},{"Start":"06:54.535 ","End":"07:00.040","Text":"The molar mass of water is 18 grams, 18.0 grams."},{"Start":"07:00.040 ","End":"07:01.985","Text":"We saw how to do that before."},{"Start":"07:01.985 ","End":"07:06.135","Text":"We need the molar mass of each atom."},{"Start":"07:06.135 ","End":"07:09.695","Text":"We\u0027ve 6 carbon atoms, 12 hydrogen atoms,"},{"Start":"07:09.695 ","End":"07:14.905","Text":"6 oxygen atoms and we add them up and we get to a 180 grams."},{"Start":"07:14.905 ","End":"07:17.000","Text":"In addition, at the end,"},{"Start":"07:17.000 ","End":"07:19.665","Text":"we need to convert the mass to the volume,"},{"Start":"07:19.665 ","End":"07:22.355","Text":"and the mass of water to the volume of water."},{"Start":"07:22.355 ","End":"07:23.870","Text":"How do we do that?"},{"Start":"07:23.870 ","End":"07:26.195","Text":"We learnt this right at the beginning of the course."},{"Start":"07:26.195 ","End":"07:28.190","Text":"We use the equation for density,"},{"Start":"07:28.190 ","End":"07:31.055","Text":"that\u0027s d = m/v."},{"Start":"07:31.055 ","End":"07:34.955","Text":"The density is equal to the mass divided by the volume."},{"Start":"07:34.955 ","End":"07:37.040","Text":"But we need the volume,"},{"Start":"07:37.040 ","End":"07:39.995","Text":"so we multiply both sides by v,"},{"Start":"07:39.995 ","End":"07:43.775","Text":"we get vd = m,"},{"Start":"07:43.775 ","End":"07:52.440","Text":"and from that we get v = m/d by dividing by d. Divide both sides by d,"},{"Start":"07:52.440 ","End":"07:55.380","Text":"we get v = m/d."},{"Start":"07:55.380 ","End":"07:58.465","Text":"This is the equation we require."},{"Start":"07:58.465 ","End":"08:02.675","Text":"d of course is the density and v is the volume."},{"Start":"08:02.675 ","End":"08:05.190","Text":"We need to know what the density of water is"},{"Start":"08:05.190 ","End":"08:07.910","Text":"and this is something we should perhaps remember,"},{"Start":"08:07.910 ","End":"08:12.640","Text":"that the density of water is 1 gram per milliliter."},{"Start":"08:12.640 ","End":"08:18.755","Text":"The density of water is 1 gram per milliliter or 1 kilogram per liter."},{"Start":"08:18.755 ","End":"08:22.610","Text":"In other words, if you carry a 1 liter bottle of water,"},{"Start":"08:22.610 ","End":"08:26.285","Text":"you are carrying 1 kilogram of water."},{"Start":"08:26.285 ","End":"08:29.030","Text":"Let\u0027s start to solve the problem."},{"Start":"08:29.030 ","End":"08:34.130","Text":"The first step, the step 1 is the mass of glucose to the moles of glucose."},{"Start":"08:34.130 ","End":"08:39.004","Text":"We have to convert the mass of glucose to the moles of glucose."},{"Start":"08:39.004 ","End":"08:43.100","Text":"The number of moles of glucose is 270 grams."},{"Start":"08:43.100 ","End":"08:47.490","Text":"That\u0027s the mass of glucose divided by its molar mass,"},{"Start":"08:47.490 ","End":"08:49.925","Text":"180 grams per mole."},{"Start":"08:49.925 ","End":"08:51.435","Text":"We divide the 2,"},{"Start":"08:51.435 ","End":"08:54.405","Text":"we get 1.5 moles."},{"Start":"08:54.405 ","End":"08:58.910","Text":"We have 1.5 moles of glucose."},{"Start":"08:58.910 ","End":"09:00.680","Text":"Now we have Step 2,"},{"Start":"09:00.680 ","End":"09:02.390","Text":"which is the central step."},{"Start":"09:02.390 ","End":"09:06.935","Text":"We have to convert the moles of glucose to moles of water."},{"Start":"09:06.935 ","End":"09:09.635","Text":"The number of moles of water,"},{"Start":"09:09.635 ","End":"09:11.815","Text":"that\u0027s what we need to find out."},{"Start":"09:11.815 ","End":"09:13.670","Text":"After writing moles of water,"},{"Start":"09:13.670 ","End":"09:15.290","Text":"you don\u0027t really need to write it,"},{"Start":"09:15.290 ","End":"09:17.540","Text":"but I want to be very clear,"},{"Start":"09:17.540 ","End":"09:18.995","Text":"at least at the beginning."},{"Start":"09:18.995 ","End":"09:22.490","Text":"Number of moles of water and the units are, of course,"},{"Start":"09:22.490 ","End":"09:28.335","Text":"moles of water is equal the number of moles of glucose,"},{"Start":"09:28.335 ","End":"09:30.595","Text":"and I\u0027ve written moles of glucose,"},{"Start":"09:30.595 ","End":"09:33.125","Text":"times our conversion factor,"},{"Start":"09:33.125 ","End":"09:36.035","Text":"the one that relates water to glucose,"},{"Start":"09:36.035 ","End":"09:40.160","Text":"so 6 moles of water is equivalent to 1 mole of glucose."},{"Start":"09:40.160 ","End":"09:44.410","Text":"Here it is. Let\u0027s solve the numbers first."},{"Start":"09:44.410 ","End":"09:49.365","Text":"We know we have 1.5 moles of glucose,"},{"Start":"09:49.365 ","End":"09:51.750","Text":"we can write 1.5,"},{"Start":"09:51.750 ","End":"09:55.405","Text":"and here we have 6 divided by 1, so 6."},{"Start":"09:55.405 ","End":"09:56.840","Text":"What about the units?"},{"Start":"09:56.840 ","End":"09:58.490","Text":"We have moles of glucose on the top,"},{"Start":"09:58.490 ","End":"10:00.895","Text":"moles of glucose in the bottom."},{"Start":"10:00.895 ","End":"10:04.215","Text":"We\u0027re left with moles of water,"},{"Start":"10:04.215 ","End":"10:08.250","Text":"1.5 times 6 is 9."},{"Start":"10:08.250 ","End":"10:12.420","Text":"Here\u0027s our answer, 9 moles of water."},{"Start":"10:12.420 ","End":"10:14.725","Text":"Now here\u0027s Step 3."},{"Start":"10:14.725 ","End":"10:18.295","Text":"We need to convert the moles of water to the mass of water."},{"Start":"10:18.295 ","End":"10:24.155","Text":"Our equation is, the mass is equal to the number of moles times the molar mass."},{"Start":"10:24.155 ","End":"10:26.285","Text":"We have 9 moles of water."},{"Start":"10:26.285 ","End":"10:30.400","Text":"The molar mass of water is 18.0 grams per mole."},{"Start":"10:30.400 ","End":"10:35.685","Text":"We multiply 9 times 18, we get 162."},{"Start":"10:35.685 ","End":"10:38.970","Text":"Moles cancels with mole minus 1."},{"Start":"10:38.970 ","End":"10:41.940","Text":"Moles times moles minus 1 is just 1."},{"Start":"10:41.940 ","End":"10:44.114","Text":"We\u0027re left just with a gram."},{"Start":"10:44.114 ","End":"10:47.510","Text":"We have a 162 grams."},{"Start":"10:47.510 ","End":"10:50.584","Text":"Now we have the mass of the water produced."},{"Start":"10:50.584 ","End":"10:54.505","Text":"Here we have the final step, Step 4."},{"Start":"10:54.505 ","End":"10:58.040","Text":"The mass of water to the volume of water."},{"Start":"10:58.040 ","End":"11:03.185","Text":"We know from above that the volume\u0027s equal to the mass divide by density."},{"Start":"11:03.185 ","End":"11:05.540","Text":"The mass is 162 grams,"},{"Start":"11:05.540 ","End":"11:07.070","Text":"as we saw up here,"},{"Start":"11:07.070 ","End":"11:08.900","Text":"divided by the density,"},{"Start":"11:08.900 ","End":"11:11.665","Text":"which is 1 gram per milliliter,"},{"Start":"11:11.665 ","End":"11:15.780","Text":"so 162 divided by 1 is just a 162."},{"Start":"11:15.780 ","End":"11:22.725","Text":"Grams cancels with grams and we\u0027re leftover 1 over milliliter to minus 1,"},{"Start":"11:22.725 ","End":"11:24.745","Text":"which is just milliliter."},{"Start":"11:24.745 ","End":"11:30.470","Text":"Here we have our final answer, 162 milliliters."},{"Start":"11:30.470 ","End":"11:32.645","Text":"That\u0027s the answer to the problem."},{"Start":"11:32.645 ","End":"11:35.690","Text":"In this video, we learned about the connection"},{"Start":"11:35.690 ","End":"11:39.005","Text":"between chemical equations and stoichiometry."},{"Start":"11:39.005 ","End":"11:44.245","Text":"We learned how to use stoichiometry to solve numerical problems."},{"Start":"11:44.245 ","End":"11:49.350","Text":"There will be many more examples in the exercises."}],"ID":17668},{"Watched":false,"Name":"Exercise 3","Duration":"1m 30s","ChapterTopicVideoID":22861,"CourseChapterTopicPlaylistID":86817,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22861.jpeg","UploadDate":"2020-12-15T05:57:55.7230000","DurationForVideoObject":"PT1M30S","Description":null,"VideoComments":[],"Subtitles":[],"ID":23682},{"Watched":false,"Name":"Exercise 4","Duration":"7m 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39s","ChapterTopicVideoID":22869,"CourseChapterTopicPlaylistID":86817,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22869.jpeg","UploadDate":"2020-12-15T06:00:06.2100000","DurationForVideoObject":"PT3M39S","Description":null,"VideoComments":[],"Subtitles":[],"ID":23690},{"Watched":false,"Name":"Limiting Reactants","Duration":"10m 49s","ChapterTopicVideoID":16924,"CourseChapterTopicPlaylistID":86817,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/16924.jpeg","UploadDate":"2019-02-20T23:51:15.3370000","DurationForVideoObject":"PT10M49S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.530","Text":"In a previous video,"},{"Start":"00:02.530 ","End":"00:06.330","Text":"we balanced the chemical equation for the combustion of glucose."},{"Start":"00:06.330 ","End":"00:11.010","Text":"We found that 1 mole of glucose reacts with 6 moles of oxygen"},{"Start":"00:11.010 ","End":"00:16.409","Text":"to give 6 moles of carbon dioxide and 6 moles of water."},{"Start":"00:16.409 ","End":"00:21.300","Text":"Now we can make this equation a little bit more precise by adding some notation."},{"Start":"00:21.300 ","End":"00:24.705","Text":"We can add s for solid, l for liquid,"},{"Start":"00:24.705 ","End":"00:28.995","Text":"g for gas, and aq for if it\u0027s dissolved in water."},{"Start":"00:28.995 ","End":"00:32.265","Text":"That\u0027s from Aqua which is a Latin for water."},{"Start":"00:32.265 ","End":"00:38.060","Text":"What we get is glucose which is a solid reacting with oxygen which is"},{"Start":"00:38.060 ","End":"00:43.940","Text":"a gas to give carbon dioxide which is a gas and water which is a liquid."},{"Start":"00:43.940 ","End":"00:48.800","Text":"Or if it\u0027s very high temperature as it probably is when you burn something,"},{"Start":"00:48.800 ","End":"00:50.995","Text":"it could be a gas."},{"Start":"00:50.995 ","End":"00:56.690","Text":"Often we write the conditions under which the reaction is performed above the arrow."},{"Start":"00:56.690 ","End":"01:01.280","Text":"For example, this Delta is capital"},{"Start":"01:01.280 ","End":"01:06.620","Text":"Delta means that it is performed at very high temperatures."},{"Start":"01:06.620 ","End":"01:11.530","Text":"Now in a previous video we also talked about stoichiometry."},{"Start":"01:11.530 ","End":"01:14.925","Text":"Remember chemical bookkeeping."},{"Start":"01:14.925 ","End":"01:23.110","Text":"These numbers, the number of moles tell us precisely how much of each component we need."},{"Start":"01:23.110 ","End":"01:26.935","Text":"1 mole of glucose will react with 6 moles of"},{"Start":"01:26.935 ","End":"01:32.800","Text":"oxygen and give us 6 moles of carbon dioxide and 6 moles of water."},{"Start":"01:32.800 ","End":"01:36.010","Text":"Now sometimes we don\u0027t have enough of one of"},{"Start":"01:36.010 ","End":"01:39.160","Text":"the reactants for the reaction to go to completion."},{"Start":"01:39.160 ","End":"01:43.825","Text":"That means that some of the reactants are left when the reaction is finished."},{"Start":"01:43.825 ","End":"01:46.690","Text":"Now the substance, the component,"},{"Start":"01:46.690 ","End":"01:50.260","Text":"the reactant that prevents the reaction from going to"},{"Start":"01:50.260 ","End":"01:54.100","Text":"completion is called a limiting reactant."},{"Start":"01:54.100 ","End":"01:55.915","Text":"Let\u0027s take some examples."},{"Start":"01:55.915 ","End":"02:00.559","Text":"Example 1, supposing we want to burn 1 mole of glucose,"},{"Start":"02:00.559 ","End":"02:04.100","Text":"but only have 5 moles of oxygen available,"},{"Start":"02:04.100 ","End":"02:08.905","Text":"how much glucose will remain after all the oxygen is consumed?"},{"Start":"02:08.905 ","End":"02:11.840","Text":"Now when we wrote the chemical equation,"},{"Start":"02:11.840 ","End":"02:17.105","Text":"we saw that we need 6 moles of oxygen to completely burn 1 mole of glucose."},{"Start":"02:17.105 ","End":"02:19.365","Text":"This problem we already have 5."},{"Start":"02:19.365 ","End":"02:23.930","Text":"Obviously we don\u0027t have enough to burn all the glucose."},{"Start":"02:23.930 ","End":"02:26.615","Text":"How much glucose will we actually burn?"},{"Start":"02:26.615 ","End":"02:30.880","Text":"We first need the conversion fact or the stoichiometric ratio."},{"Start":"02:30.880 ","End":"02:35.285","Text":"That\u0027s 1 mole of glucose is equivalent to 6 moles of oxygen."},{"Start":"02:35.285 ","End":"02:39.785","Text":"1 mole of glucose requires 6 moles of oxygen to burn."},{"Start":"02:39.785 ","End":"02:43.195","Text":"We can write this ratio and it\u0027s equal to 1."},{"Start":"02:43.195 ","End":"02:45.680","Text":"That\u0027s our stoichiometric ratio."},{"Start":"02:45.680 ","End":"02:47.280","Text":"Now here\u0027s the calculation."},{"Start":"02:47.280 ","End":"02:49.980","Text":"The number of moles of glucose."},{"Start":"02:49.980 ","End":"02:54.210","Text":"I\u0027ve written out in total and full."},{"Start":"02:54.210 ","End":"02:56.284","Text":"Number of moles of glucose,"},{"Start":"02:56.284 ","End":"02:59.869","Text":"moles of glucose is equal to the number of moles of oxygen,"},{"Start":"02:59.869 ","End":"03:02.420","Text":"moles of oxygen that\u0027s units."},{"Start":"03:02.420 ","End":"03:05.760","Text":"Here\u0027s our stoichiometric ratio."},{"Start":"03:05.760 ","End":"03:10.250","Text":"1 mole of glucose divided by 6 moles of oxygen."},{"Start":"03:10.250 ","End":"03:13.880","Text":"Now we can put in the numbers and oxygen,"},{"Start":"03:13.880 ","End":"03:16.630","Text":"number of moles of oxygen is 5."},{"Start":"03:16.630 ","End":"03:22.769","Text":"Here\u0027s the 6, so it\u0027s 5 divided by 6 5/6ths and we need the units."},{"Start":"03:22.769 ","End":"03:26.040","Text":"We have moles of oxygen on the numerator,"},{"Start":"03:26.040 ","End":"03:28.665","Text":"moles of oxygen in the denominator,"},{"Start":"03:28.665 ","End":"03:31.440","Text":"and we\u0027re left with moles of glucose."},{"Start":"03:31.440 ","End":"03:34.905","Text":"We have 5/6 moles of glucose."},{"Start":"03:34.905 ","End":"03:39.800","Text":"From that, we can conclude that only 5/6 moles of"},{"Start":"03:39.800 ","End":"03:44.930","Text":"glucose will be used up and 1/6 of the mole will be left."},{"Start":"03:44.930 ","End":"03:47.965","Text":"Because we started off with 1 whole mole."},{"Start":"03:47.965 ","End":"03:49.965","Text":"We used 5/6th,"},{"Start":"03:49.965 ","End":"03:52.115","Text":"so 1/6th will be left."},{"Start":"03:52.115 ","End":"03:54.515","Text":"What could we say about that?"},{"Start":"03:54.515 ","End":"03:58.385","Text":"We can say that oxygen is the limiting reactant."},{"Start":"03:58.385 ","End":"04:03.265","Text":"Oxygen is what prevented us from burning all the glucose."},{"Start":"04:03.265 ","End":"04:05.615","Text":"Since we had glucose leftover,"},{"Start":"04:05.615 ","End":"04:08.405","Text":"glucose is in excess."},{"Start":"04:08.405 ","End":"04:10.820","Text":"Let\u0027s take a more complicated example."},{"Start":"04:10.820 ","End":"04:12.230","Text":"Example 2."},{"Start":"04:12.230 ","End":"04:16.190","Text":"If 88.3 grams of sulfuric acid reacts with"},{"Start":"04:16.190 ","End":"04:22.385","Text":"68.0 grams of sodium hydroxide to form sodium sulfate and water,"},{"Start":"04:22.385 ","End":"04:24.995","Text":"which is the limiting reactant?"},{"Start":"04:24.995 ","End":"04:28.774","Text":"The first thing we need to do is to write the equation."},{"Start":"04:28.774 ","End":"04:31.535","Text":"Now what reaction is this?"},{"Start":"04:31.535 ","End":"04:34.340","Text":"This reaction is a reaction between"},{"Start":"04:34.340 ","End":"04:40.365","Text":"a strong acid and a strong base sodium hydroxide base."},{"Start":"04:40.365 ","End":"04:44.825","Text":"It\u0027s going to give us a salt plus water."},{"Start":"04:44.825 ","End":"04:49.280","Text":"The first step is to write the equation and balance it."},{"Start":"04:49.280 ","End":"04:51.005","Text":"I\u0027m going to write the equation."},{"Start":"04:51.005 ","End":"04:54.110","Text":"You don\u0027t yet know how to write such an equation."},{"Start":"04:54.110 ","End":"04:55.840","Text":"I\u0027m going to write it for you."},{"Start":"04:55.840 ","End":"04:58.540","Text":"We\u0027ll learn more about it later."},{"Start":"04:58.540 ","End":"05:03.800","Text":"Here\u0027s sulfuric acid H_2SO_4 and it\u0027s in water,"},{"Start":"05:03.800 ","End":"05:07.370","Text":"so it\u0027s aqueous plus NaOH,"},{"Start":"05:07.370 ","End":"05:11.765","Text":"that\u0027s sodium hydroxide and it\u0027s also in water aq,"},{"Start":"05:11.765 ","End":"05:15.065","Text":"to give sodium sulfate."},{"Start":"05:15.065 ","End":"05:18.500","Text":"That\u0027s also in water because it\u0027s soluble in water."},{"Start":"05:18.500 ","End":"05:23.735","Text":"Here\u0027s the water. We get a little bit extra water, L for liquid."},{"Start":"05:23.735 ","End":"05:26.890","Text":"Now we need to balance this equation."},{"Start":"05:26.890 ","End":"05:30.270","Text":"If we look at the equation we see"},{"Start":"05:30.270 ","End":"05:38.460","Text":"that there\u0027s only 1 sodium on the left hand side and 2 on the right hand side."},{"Start":"05:38.460 ","End":"05:42.130","Text":"We need to multiply Na by 2,"},{"Start":"05:42.130 ","End":"05:46.435","Text":"we can only multiply the whole of NaOH."},{"Start":"05:46.435 ","End":"05:48.615","Text":"Let\u0027s look at the hydrogens."},{"Start":"05:48.615 ","End":"05:50.460","Text":"2 hydrogens here."},{"Start":"05:50.460 ","End":"05:53.565","Text":"Another 2 hydrogens here 2 times 1."},{"Start":"05:53.565 ","End":"05:55.685","Text":"That\u0027s 4 altogether."},{"Start":"05:55.685 ","End":"06:01.920","Text":"Here we have just 2 H_2O, just 2 hydrogens."},{"Start":"06:01.920 ","End":"06:06.525","Text":"We need to multiply water by 2 to make 4 hydrogens."},{"Start":"06:06.525 ","End":"06:09.416","Text":"Now the equation is balanced."},{"Start":"06:09.416 ","End":"06:18.340","Text":"H_2SO_4 plus 2NaOH to give Na_2SO_4 plus 2H2O."},{"Start":"06:18.340 ","End":"06:20.951","Text":"Now we can write the number of moles."},{"Start":"06:20.951 ","End":"06:24.740","Text":"1 mole of sulfuric acid,"},{"Start":"06:24.740 ","End":"06:31.760","Text":"2 moles of sodium hydroxide to give 1 mole of sodium sulfate and 2 moles of water."},{"Start":"06:31.760 ","End":"06:34.940","Text":"The second step is to calculate the number of moles"},{"Start":"06:34.940 ","End":"06:38.555","Text":"of each substance using the molar masses."},{"Start":"06:38.555 ","End":"06:45.295","Text":"The molar mass of sulfuric acid we can work it out is 98.1 grams per mole."},{"Start":"06:45.295 ","End":"06:49.985","Text":"The molar mass of sodium hydroxide is 40.0 grams per mole."},{"Start":"06:49.985 ","End":"06:53.330","Text":"Now the number of moles of sulfuric acid is the mass of"},{"Start":"06:53.330 ","End":"06:58.165","Text":"sulfuric acid which is 88.3 grams we\u0027re given that in the question,"},{"Start":"06:58.165 ","End":"07:03.570","Text":"divided by the molar mass 98.1 grams per mole,"},{"Start":"07:03.570 ","End":"07:07.230","Text":"that gives us 0.90 moles."},{"Start":"07:07.230 ","End":"07:14.240","Text":"The number of moles of sodium hydroxide is the mass of sodium hydroxide 68.0 grams"},{"Start":"07:14.240 ","End":"07:22.015","Text":"divided by the molar mass 40.0 grams per mole and that gives us 1.7 moles."},{"Start":"07:22.015 ","End":"07:26.780","Text":"Step 3 is to check which reactant is the limiting one."},{"Start":"07:26.780 ","End":"07:33.440","Text":"Let\u0027s look at the situation that we have 0.90"},{"Start":"07:33.440 ","End":"07:39.935","Text":"moles of sulfuric acid and see how much sodium hydroxide reacts."},{"Start":"07:39.935 ","End":"07:47.840","Text":"Or we could have 1.7 moles of sodium hydroxide and see how much sulfuric acid reacts."},{"Start":"07:47.840 ","End":"07:51.080","Text":"To do this, we first need the conversion factor"},{"Start":"07:51.080 ","End":"07:54.200","Text":"for the reactants from the chemical equation."},{"Start":"07:54.200 ","End":"07:59.780","Text":"It\u0027s 1 mole sulfuric acid to 2 moles of sodium hydroxide."},{"Start":"07:59.780 ","End":"08:01.680","Text":"That\u0027s our conversion factor."},{"Start":"08:01.680 ","End":"08:09.590","Text":"Now if the number of moles of sulfuric acid is 0.90 moles then the number of moles of"},{"Start":"08:09.590 ","End":"08:13.400","Text":"sodium hydroxide we require is 0.90"},{"Start":"08:13.400 ","End":"08:18.145","Text":"moles of sulfuric acid multiplied by the conversion factor,"},{"Start":"08:18.145 ","End":"08:22.940","Text":"2 moles of sodium hydroxide to 1 mole of sulfuric acid."},{"Start":"08:22.940 ","End":"08:25.100","Text":"We can work out the numbers first,"},{"Start":"08:25.100 ","End":"08:29.900","Text":"0.90 times 2 gives us 1.8."},{"Start":"08:29.900 ","End":"08:36.215","Text":"Then the units moles of sulfuric acid cancels with moles of sulfuric acid,"},{"Start":"08:36.215 ","End":"08:40.190","Text":"leaves us moles of sodium hydroxide and the answer we"},{"Start":"08:40.190 ","End":"08:45.060","Text":"obtain is 1.8 moles of sodium hydroxide."},{"Start":"08:45.060 ","End":"08:49.120","Text":"Now let\u0027s consider the second possibility."},{"Start":"08:50.750 ","End":"08:55.824","Text":"Supposing we have 1.7 moles of sodium hydroxide,"},{"Start":"08:55.824 ","End":"08:59.950","Text":"then how many moles of sulfuric acid will react?"},{"Start":"08:59.950 ","End":"09:03.530","Text":"The moles of sulfuric acid is 1.7 moles of"},{"Start":"09:03.530 ","End":"09:07.396","Text":"sodium hydroxide multiplied by our conversion factor,"},{"Start":"09:07.396 ","End":"09:12.244","Text":"1 mole of sulfuric acid to 2 moles of sodium hydroxide."},{"Start":"09:12.244 ","End":"09:19.770","Text":"Let\u0027s take the numbers 1.7 divided by 2 is 0.85 and then the units."},{"Start":"09:19.770 ","End":"09:23.900","Text":"Moles of sodium hydroxide cancels with moles of"},{"Start":"09:23.900 ","End":"09:28.750","Text":"sodium hydroxide to give us moles of sulfuric acid."},{"Start":"09:28.750 ","End":"09:33.050","Text":"We have 0.85 moles of sulfuric acid."},{"Start":"09:33.050 ","End":"09:40.925","Text":"Now for all the sulfuric acid to react we need 1.8 moles of sodium hydroxide."},{"Start":"09:40.925 ","End":"09:42.470","Text":"That\u0027s what we have here."},{"Start":"09:42.470 ","End":"09:45.740","Text":"But we only have 1.7 moles."},{"Start":"09:45.740 ","End":"09:47.825","Text":"That\u0027s what we calculated before."},{"Start":"09:47.825 ","End":"09:50.120","Text":"We only have 1.7."},{"Start":"09:50.120 ","End":"09:55.355","Text":"We don\u0027t have enough for all the sulfuric acid to react."},{"Start":"09:55.355 ","End":"10:00.455","Text":"Therefore sodium hydroxide is the limiting reactant."},{"Start":"10:00.455 ","End":"10:02.900","Text":"We can look at it another way."},{"Start":"10:02.900 ","End":"10:11.530","Text":"We need 0.85 moles of sulfuric acid for all the NaOH to react."},{"Start":"10:11.530 ","End":"10:14.535","Text":"We have 0.9 moles."},{"Start":"10:14.535 ","End":"10:18.155","Text":"There\u0027ll be sulfuric acid left after the reaction."},{"Start":"10:18.155 ","End":"10:22.590","Text":"It will be in excess. We have 0.9."},{"Start":"10:22.590 ","End":"10:25.860","Text":"We only require 0.85."},{"Start":"10:25.860 ","End":"10:32.715","Text":"We can calculate how much will be left over 0.05 moles."},{"Start":"10:32.715 ","End":"10:36.320","Text":"In this video we saw that if the reactants are not in"},{"Start":"10:36.320 ","End":"10:42.005","Text":"the stoichiometric ratio that was defined by the chemical equation,"},{"Start":"10:42.005 ","End":"10:49.020","Text":"one reactant will be the limiting reactant and the other will be in excess."}],"ID":17669},{"Watched":false,"Name":"Stoichiometry in Solutions","Duration":"13m 31s","ChapterTopicVideoID":16925,"CourseChapterTopicPlaylistID":86817,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/16925.jpeg","UploadDate":"2019-02-20T23:52:46.9630000","DurationForVideoObject":"PT13M31S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.170 ","End":"00:02.565","Text":"In a previous video,"},{"Start":"00:02.565 ","End":"00:05.910","Text":"we introduced the concept of stoichiometry."},{"Start":"00:05.910 ","End":"00:10.620","Text":"In this video, we will discuss stoichiometry in solutions."},{"Start":"00:10.620 ","End":"00:11.940","Text":"In order to begin,"},{"Start":"00:11.940 ","End":"00:13.785","Text":"we need some definitions."},{"Start":"00:13.785 ","End":"00:15.915","Text":"We need to define solute,"},{"Start":"00:15.915 ","End":"00:17.880","Text":"solvent, and solution."},{"Start":"00:17.880 ","End":"00:21.375","Text":"Let\u0027s begin with solvent."},{"Start":"00:21.375 ","End":"00:26.415","Text":"The solvent is the substance in which the substance of interest is dissolved."},{"Start":"00:26.415 ","End":"00:28.170","Text":"It could be, for example,"},{"Start":"00:28.170 ","End":"00:31.215","Text":"as it will be in the rest of this video, water."},{"Start":"00:31.215 ","End":"00:37.020","Text":"Or it could be some organic solvent used to clean your clothes."},{"Start":"00:37.020 ","End":"00:40.655","Text":"The solute is a substance dissolved in the solvent."},{"Start":"00:40.655 ","End":"00:45.080","Text":"For example, if I dissolve salt in water,"},{"Start":"00:45.080 ","End":"00:47.090","Text":"salt will be the solute,"},{"Start":"00:47.090 ","End":"00:49.430","Text":"and water will be the solvent."},{"Start":"00:49.430 ","End":"00:53.750","Text":"The solution is obtained when the solute is dissolved in the solvent."},{"Start":"00:53.750 ","End":"00:59.045","Text":"For example, salty water is a solution of salt in water."},{"Start":"00:59.045 ","End":"01:03.395","Text":"In this video, we will talk mainly about aqueous solutions,"},{"Start":"01:03.395 ","End":"01:07.415","Text":"solutions in water, but there are many other solutions."},{"Start":"01:07.415 ","End":"01:13.700","Text":"For example, air is a solution where the solvent is nitrogen,"},{"Start":"01:13.700 ","End":"01:18.250","Text":"and the solute is oxygen,"},{"Start":"01:18.250 ","End":"01:21.785","Text":"or carbon dioxide, or a few other things."},{"Start":"01:21.785 ","End":"01:24.905","Text":"Or another example is soda."},{"Start":"01:24.905 ","End":"01:26.960","Text":"Soda, which we drink."},{"Start":"01:26.960 ","End":"01:32.090","Text":"Soda is a solution of carbon dioxide in water."},{"Start":"01:32.090 ","End":"01:35.420","Text":"Now let\u0027s talk about aqueous solutions."},{"Start":"01:35.420 ","End":"01:40.060","Text":"We\u0027re going to take as an example a solution of sodium chloride in water."},{"Start":"01:40.060 ","End":"01:45.235","Text":"Here sodium chloride is the solute and water is the solvent,"},{"Start":"01:45.235 ","End":"01:50.495","Text":"we write it as NaCl with aq in brackets after it,"},{"Start":"01:50.495 ","End":"01:53.885","Text":"that\u0027s indicating an aqueous solution."},{"Start":"01:53.885 ","End":"01:56.780","Text":"Now we\u0027re going to talk about concentration."},{"Start":"01:56.780 ","End":"02:00.845","Text":"We can have solutions of varying concentrations."},{"Start":"02:00.845 ","End":"02:04.010","Text":"We could have a very salty solution that would be"},{"Start":"02:04.010 ","End":"02:08.495","Text":"a higher concentration than the barely salty solution."},{"Start":"02:08.495 ","End":"02:14.555","Text":"Now there are many ways of defining concentration."},{"Start":"02:14.555 ","End":"02:19.175","Text":"The one we\u0027re going to talk about this video is called molarity."},{"Start":"02:19.175 ","End":"02:25.579","Text":"Rewrite it as capital M. The molarity of a solution is defined as follows."},{"Start":"02:25.579 ","End":"02:34.760","Text":"The molarity is equal to n where it\u0027s the number of moles of solute divided by V,"},{"Start":"02:34.760 ","End":"02:38.495","Text":"that\u0027s the volume of the solution in liters."},{"Start":"02:38.495 ","End":"02:42.845","Text":"The volume must be in liters, not in milliliters."},{"Start":"02:42.845 ","End":"02:46.400","Text":"How can we prepare an aqueous solution?"},{"Start":"02:46.400 ","End":"02:50.824","Text":"In this picture, we have a volumetric flask."},{"Start":"02:50.824 ","End":"02:52.745","Text":"This is this flask."},{"Start":"02:52.745 ","End":"02:56.450","Text":"It has a mark on it giving a particular volume,"},{"Start":"02:56.450 ","End":"02:59.270","Text":"here it\u0027s 250 milliliters."},{"Start":"02:59.270 ","End":"03:02.135","Text":"We put the solid we\u0027re interested in,"},{"Start":"03:02.135 ","End":"03:04.640","Text":"in the base of the flask."},{"Start":"03:04.640 ","End":"03:05.990","Text":"Usually we weigh it,"},{"Start":"03:05.990 ","End":"03:07.760","Text":"and then put it there."},{"Start":"03:07.760 ","End":"03:14.270","Text":"Then we add a little of the solvent sufficient to dissolve the solute,"},{"Start":"03:14.270 ","End":"03:16.100","Text":"and then finally,"},{"Start":"03:16.100 ","End":"03:25.235","Text":"we add enough solvent to reach the mark on the neck of the flask."},{"Start":"03:25.235 ","End":"03:29.870","Text":"This flask is called a volumetric flask."},{"Start":"03:29.870 ","End":"03:33.950","Text":"We can find them in all different volumes."},{"Start":"03:33.950 ","End":"03:37.895","Text":"Let\u0027s take an example based on this picture."},{"Start":"03:37.895 ","End":"03:42.260","Text":"What is the molarity of an aqueous solution of NaCl when"},{"Start":"03:42.260 ","End":"03:49.300","Text":"0.584 grams is dissolved in 250 milliliters of water?"},{"Start":"03:49.300 ","End":"03:52.715","Text":"In order to calculate the number of moles,"},{"Start":"03:52.715 ","End":"03:57.410","Text":"we first need to know what the molar mass is of NaCl."},{"Start":"03:57.410 ","End":"04:03.005","Text":"The molar mass of NaCl is 58.4 grams per mole."},{"Start":"04:03.005 ","End":"04:09.470","Text":"We get it by adding up the molar mass of sodium to the molar mass of chlorine."},{"Start":"04:09.470 ","End":"04:14.765","Text":"Now the number of moles of NaCl is its mass divided by molar mass."},{"Start":"04:14.765 ","End":"04:17.885","Text":"It\u0027s mass, according to the question we\u0027ve been asked,"},{"Start":"04:17.885 ","End":"04:20.705","Text":"is 0.584 grams,"},{"Start":"04:20.705 ","End":"04:24.785","Text":"and its molar mass is 58.4 grams per mole."},{"Start":"04:24.785 ","End":"04:26.955","Text":"When we divide these 2,"},{"Start":"04:26.955 ","End":"04:29.030","Text":"let\u0027s deal with the numbers first of all,"},{"Start":"04:29.030 ","End":"04:31.240","Text":"we get 0.01,"},{"Start":"04:31.240 ","End":"04:32.580","Text":"and then the units,"},{"Start":"04:32.580 ","End":"04:35.055","Text":"the grams cancels with grams."},{"Start":"04:35.055 ","End":"04:37.950","Text":"1 over molar to the minus 1 is mole,"},{"Start":"04:37.950 ","End":"04:42.525","Text":"so we get 0.01 moles of NaCl."},{"Start":"04:42.525 ","End":"04:45.135","Text":"Now we need the volume."},{"Start":"04:45.135 ","End":"04:48.710","Text":"We were told that the volume is 250 milliliters,"},{"Start":"04:48.710 ","End":"04:51.745","Text":"but we need the volume in liters."},{"Start":"04:51.745 ","End":"04:55.010","Text":"We multiply it by this conversion factor,"},{"Start":"04:55.010 ","End":"04:59.030","Text":"1 liter is the same as 1,000 milliliters,"},{"Start":"04:59.030 ","End":"05:04.100","Text":"250 divided by 1,000 is 0.254."},{"Start":"05:04.100 ","End":"05:08.920","Text":"For units the milliliters here cancels with milliliters here,"},{"Start":"05:08.920 ","End":"05:11.100","Text":"and we are left with liters,"},{"Start":"05:11.100 ","End":"05:14.960","Text":"so the volume is 0.25 liters."},{"Start":"05:14.960 ","End":"05:17.075","Text":"Now we can calculate the molarity."},{"Start":"05:17.075 ","End":"05:21.155","Text":"Molarity is the number of moles, 0.01 moles,"},{"Start":"05:21.155 ","End":"05:27.605","Text":"divided by the volume of the solution in liters, 0.25 liters."},{"Start":"05:27.605 ","End":"05:31.240","Text":"Divide the numbers we get 0.04."},{"Start":"05:31.240 ","End":"05:34.470","Text":"The units are moles per liter."},{"Start":"05:34.870 ","End":"05:40.960","Text":"We can say the molarity is 0.04M."},{"Start":"05:40.960 ","End":"05:43.275","Text":"That\u0027s the molarity,"},{"Start":"05:43.275 ","End":"05:48.020","Text":"that\u0027s the concentration of the salt in water."},{"Start":"05:48.020 ","End":"05:53.750","Text":"Now sometimes we need to perform the opposite of what we\u0027ve just done."},{"Start":"05:53.750 ","End":"05:59.200","Text":"We need to calculate the number of moles from the volume, and concentration."},{"Start":"05:59.200 ","End":"06:01.720","Text":"Let\u0027s take the equation for the molarity."},{"Start":"06:01.720 ","End":"06:05.110","Text":"The Molarity is number of moles divided by the volume."},{"Start":"06:05.110 ","End":"06:09.100","Text":"If we multiply both sides of this equation by V,"},{"Start":"06:09.100 ","End":"06:13.740","Text":"we get on 1 side m times V,"},{"Start":"06:13.740 ","End":"06:17.640","Text":"that\u0027s here, and the other side we get n,"},{"Start":"06:17.640 ","End":"06:21.135","Text":"we have n equals M times V."},{"Start":"06:21.135 ","End":"06:25.644","Text":"The number of moles is equal to the molarity times the volume."},{"Start":"06:25.644 ","End":"06:27.235","Text":"Let\u0027s take an example."},{"Start":"06:27.235 ","End":"06:30.655","Text":"How many moles of NaCl are required to prepare"},{"Start":"06:30.655 ","End":"06:36.350","Text":"100 milliliters of 1 molar aqueous solution?"},{"Start":"06:36.350 ","End":"06:42.095","Text":"Now, we know that the number of moles is equal to the molarity times the volume."},{"Start":"06:42.095 ","End":"06:44.010","Text":"That\u0027s 1 M,"},{"Start":"06:44.010 ","End":"06:48.965","Text":"that\u0027s the molarity times the volume 0.1 liters."},{"Start":"06:48.965 ","End":"06:52.280","Text":"Now we can write out the units in full,"},{"Start":"06:52.280 ","End":"06:56.405","Text":"M is moles per liter."},{"Start":"06:56.405 ","End":"06:59.750","Text":"We have moles per liter times liter,"},{"Start":"06:59.750 ","End":"07:01.655","Text":"that just leaves moles,"},{"Start":"07:01.655 ","End":"07:08.885","Text":"and here\u0027s the moles, and 1 times 0.1 is of course just 0.1."},{"Start":"07:08.885 ","End":"07:13.414","Text":"The answer is 0.1 moles."},{"Start":"07:13.414 ","End":"07:18.875","Text":"Now we\u0027re going to use what we\u0027ve just calculated in a chemical reaction."},{"Start":"07:18.875 ","End":"07:23.090","Text":"The chemical reaction we have molarity, and stoichiometry."},{"Start":"07:23.090 ","End":"07:25.925","Text":"We\u0027re going to connect all these 3 concepts."},{"Start":"07:25.925 ","End":"07:29.300","Text":"Here\u0027s a reaction we considered in the previous video."},{"Start":"07:29.300 ","End":"07:36.125","Text":"Sulfuric acid reacts with sodium hydroxide to get sodium sulfate, and water."},{"Start":"07:36.125 ","End":"07:38.810","Text":"The reaction is already balanced."},{"Start":"07:38.810 ","End":"07:40.640","Text":"Here are the number of moles."},{"Start":"07:40.640 ","End":"07:43.370","Text":"1 mole of sulfuric acid,"},{"Start":"07:43.370 ","End":"07:49.295","Text":"and 2 moles of sodium hydroxide gives us 1 mole of sodium sulfate,"},{"Start":"07:49.295 ","End":"07:51.070","Text":"and 2 moles of water."},{"Start":"07:51.070 ","End":"07:57.380","Text":"Let\u0027s use an example based on this equation.100 milliliters of 0.1"},{"Start":"07:57.380 ","End":"08:05.230","Text":"molar H_2SO_4 reacts with 200 milliliters of 0.1 molar NaOH,"},{"Start":"08:05.230 ","End":"08:09.325","Text":"what is the molarity of Na_2SO_4?"},{"Start":"08:09.325 ","End":"08:14.285","Text":"We\u0027re given information about the reactants,"},{"Start":"08:14.285 ","End":"08:18.475","Text":"and we\u0027re expected to find information about the products."},{"Start":"08:18.475 ","End":"08:22.295","Text":"Here\u0027s the path we discussed in another video."},{"Start":"08:22.295 ","End":"08:27.605","Text":"The information about the reactants will lead us to calculate the moles of reactants."},{"Start":"08:27.605 ","End":"08:30.559","Text":"From that we\u0027ll calculate the moles of products."},{"Start":"08:30.559 ","End":"08:34.175","Text":"This is the central part of the problem."},{"Start":"08:34.175 ","End":"08:39.955","Text":"From the moles and products we\u0027ll find information about the products."},{"Start":"08:39.955 ","End":"08:43.620","Text":"To relate this a little better to the question we\u0027ve been asked,"},{"Start":"08:43.620 ","End":"08:45.325","Text":"we can write it like this."},{"Start":"08:45.325 ","End":"08:47.920","Text":"The concentration of volume of reactants,"},{"Start":"08:47.920 ","End":"08:50.919","Text":"that\u0027s what\u0027s given us in the question."},{"Start":"08:50.919 ","End":"08:54.400","Text":"From this we\u0027ll calculate the moles of reactants."},{"Start":"08:54.400 ","End":"08:57.670","Text":"From that we\u0027ll go to the moles of product,"},{"Start":"08:57.670 ","End":"09:01.975","Text":"and from that we\u0027ll go to the volume and concentration of the product."},{"Start":"09:01.975 ","End":"09:03.760","Text":"These are the stages,"},{"Start":"09:03.760 ","End":"09:07.445","Text":"1, 2, 3."},{"Start":"09:07.445 ","End":"09:11.355","Text":"There are 3 stages in solving this problem."},{"Start":"09:11.355 ","End":"09:17.015","Text":"In Step 1, we need to calculate the moles of sulfuric acid."},{"Start":"09:17.015 ","End":"09:23.700","Text":"The number of moles of sulfuric acid is equal to its molarity times its volume,"},{"Start":"09:23.700 ","End":"09:31.025","Text":"0.1 M times 0.1 liters."},{"Start":"09:31.025 ","End":"09:38.880","Text":"I\u0027ve already converted the 100 milliliters to 0.1 liters."},{"Start":"09:38.880 ","End":"09:43.790","Text":"If I multiply 0.1 times 0.1, I get 0.01."},{"Start":"09:43.790 ","End":"09:46.745","Text":"If I calculate capital M,"},{"Start":"09:46.745 ","End":"09:53.270","Text":"which you remember is moles per liter by liter,"},{"Start":"09:53.270 ","End":"09:55.150","Text":"I get just moles."},{"Start":"09:55.150 ","End":"09:58.515","Text":"I get 0.01 moles."},{"Start":"09:58.515 ","End":"10:03.665","Text":"Now, the number of moles of sodium hydroxide is again,"},{"Start":"10:03.665 ","End":"10:06.725","Text":"it\u0027s molarity times volume."},{"Start":"10:06.725 ","End":"10:13.010","Text":"We have 0.1 molar times 0.2 liters."},{"Start":"10:13.010 ","End":"10:18.950","Text":"Again, we get 0.02 and the units are moles."},{"Start":"10:18.950 ","End":"10:23.670","Text":"We\u0027re not going to spell it out every time."},{"Start":"10:23.670 ","End":"10:26.420","Text":"M times L,"},{"Start":"10:26.420 ","End":"10:32.374","Text":"molarity times liters will always give us moles."},{"Start":"10:32.374 ","End":"10:37.835","Text":"The moles of sulfuric acid is 0.01,"},{"Start":"10:37.835 ","End":"10:42.770","Text":"and the moles of NaOH is 0.02."},{"Start":"10:42.770 ","End":"10:49.205","Text":"Going to take a little detour to discuss whether there is a limiting reactant."},{"Start":"10:49.205 ","End":"10:52.940","Text":"Let\u0027s write the number of moles of NAOH,"},{"Start":"10:52.940 ","End":"10:58.520","Text":"that\u0027s 0.02 divided by the number of moles of sulfuric acid,"},{"Start":"10:58.520 ","End":"11:07.860","Text":"that 0.01, we divide 0.02 by 0.01 we get 2 divided by 1."},{"Start":"11:07.860 ","End":"11:12.562","Text":"That\u0027s 2 moles of NAOH divided by 1 mole of H_2SO_4,"},{"Start":"11:12.562 ","End":"11:16.955","Text":"and that\u0027s precisely the stoichiometric ratio."},{"Start":"11:16.955 ","End":"11:19.250","Text":"That\u0027s equal to 1."},{"Start":"11:19.250 ","End":"11:22.790","Text":"From all this, we can conclude that all the reactants will"},{"Start":"11:22.790 ","End":"11:27.425","Text":"react completely because they\u0027re in stoichiometric proportions."},{"Start":"11:27.425 ","End":"11:29.150","Text":"Now Step 2,"},{"Start":"11:29.150 ","End":"11:33.700","Text":"how many moles of Na_2SO_4 will be formed?"},{"Start":"11:33.700 ","End":"11:36.120","Text":"You look at the stoichiometric ratio,"},{"Start":"11:36.120 ","End":"11:42.165","Text":"1 mole of Na_2SO_4 is equivalent to 1 mole of sulfuric acid,"},{"Start":"11:42.165 ","End":"11:44.490","Text":"that ratio gives us 1."},{"Start":"11:44.490 ","End":"11:47.750","Text":"The number of moles of Na_2SO_4 will be"},{"Start":"11:47.750 ","End":"11:52.055","Text":"precisely equal to the number of moles of H_2SO_4."},{"Start":"11:52.055 ","End":"11:54.915","Text":"We found that was 0.01."},{"Start":"11:54.915 ","End":"12:00.385","Text":"The number of moles of Na_2SO_4 will be 0.01."},{"Start":"12:00.385 ","End":"12:03.290","Text":"Now, in order to complete the problem,"},{"Start":"12:03.290 ","End":"12:07.760","Text":"we need the volume and concentration of Na_2SO_4."},{"Start":"12:07.760 ","End":"12:09.455","Text":"Now in the beginning,"},{"Start":"12:09.455 ","End":"12:12.335","Text":"we had 100 milliliters of sulfuric acid,"},{"Start":"12:12.335 ","End":"12:14.945","Text":"and 200 milliliters of NAOH,"},{"Start":"12:14.945 ","End":"12:17.935","Text":"the total is 300 milliliters."},{"Start":"12:17.935 ","End":"12:23.840","Text":"The small amount of water formed in the reaction,"},{"Start":"12:23.840 ","End":"12:28.880","Text":"is really quite negligible with respect to 300 milliliters,"},{"Start":"12:28.880 ","End":"12:31.075","Text":"so we\u0027ll just ignore it,"},{"Start":"12:31.075 ","End":"12:33.985","Text":"we have 300 milliliters."},{"Start":"12:33.985 ","End":"12:38.975","Text":"Now we can calculate the molarity because we know the number of moles,"},{"Start":"12:38.975 ","End":"12:40.555","Text":"and we know the volume."},{"Start":"12:40.555 ","End":"12:43.680","Text":"The number of moles is 0.01 moles,"},{"Start":"12:43.680 ","End":"12:48.010","Text":"the volume is 0.30 liters,"},{"Start":"12:48.010 ","End":"12:53.555","Text":"I\u0027ve converted 300 milliliters to 0.30 liters."},{"Start":"12:53.555 ","End":"12:57.765","Text":"If I divide 0.01 by 0.30,"},{"Start":"12:57.765 ","End":"13:03.930","Text":"I get 0.033 in moles per liter. That\u0027s the units."},{"Start":"13:03.930 ","End":"13:07.140","Text":"Here\u0027s the answer."},{"Start":"13:07.140 ","End":"13:13.040","Text":"This tells us that the Na_2SO_4 that\u0027s formed has"},{"Start":"13:13.040 ","End":"13:19.250","Text":"a molarity of 0.033M."},{"Start":"13:19.250 ","End":"13:24.950","Text":"In this video, we"},{"Start":"13:24.950 ","End":"13:31.680","Text":"learned about stoichiometry when the reaction involves aqueous solutions."}],"ID":17670},{"Watched":false,"Name":"Exercise 7","Duration":"4m 40s","ChapterTopicVideoID":22871,"CourseChapterTopicPlaylistID":86817,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22871.jpeg","UploadDate":"2020-12-15T06:00:25.4570000","DurationForVideoObject":"PT4M40S","Description":null,"VideoComments":[],"Subtitles":[],"ID":23692},{"Watched":false,"Name":"Exercise 8","Duration":"5m 54s","ChapterTopicVideoID":22858,"CourseChapterTopicPlaylistID":86817,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22858.jpeg","UploadDate":"2020-12-15T05:56:58.2700000","DurationForVideoObject":"PT5M54S","Description":null,"VideoComments":[],"Subtitles":[],"ID":23679},{"Watched":false,"Name":"Exercise 9","Duration":"8m 11s","ChapterTopicVideoID":22860,"CourseChapterTopicPlaylistID":86817,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22860.jpeg","UploadDate":"2020-12-15T05:57:51.0130000","DurationForVideoObject":"PT8M11S","Description":null,"VideoComments":[],"Subtitles":[],"ID":23681},{"Watched":false,"Name":"Exercise 10","Duration":"3m 20s","ChapterTopicVideoID":22862,"CourseChapterTopicPlaylistID":86817,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22862.jpeg","UploadDate":"2020-12-15T05:58:05.7970000","DurationForVideoObject":"PT3M20S","Description":null,"VideoComments":[],"Subtitles":[],"ID":23683},{"Watched":false,"Name":"Exercise 11","Duration":"6m 47s","ChapterTopicVideoID":22864,"CourseChapterTopicPlaylistID":86817,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22864.jpeg","UploadDate":"2020-12-15T05:58:57.6670000","DurationForVideoObject":"PT6M47S","Description":null,"VideoComments":[],"Subtitles":[],"ID":23685},{"Watched":false,"Name":"Exercise 12","Duration":"4m 31s","ChapterTopicVideoID":22866,"CourseChapterTopicPlaylistID":86817,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22866.jpeg","UploadDate":"2020-12-15T05:59:33.0930000","DurationForVideoObject":"PT4M31S","Description":null,"VideoComments":[],"Subtitles":[],"ID":23687},{"Watched":false,"Name":"Diluting Solutions","Duration":"6m 33s","ChapterTopicVideoID":16926,"CourseChapterTopicPlaylistID":86817,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/16926.jpeg","UploadDate":"2019-02-20T23:54:02.7100000","DurationForVideoObject":"PT6M33S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.110 ","End":"00:02.430","Text":"In the previous video,"},{"Start":"00:02.430 ","End":"00:04.290","Text":"we defined the molarity."},{"Start":"00:04.290 ","End":"00:07.770","Text":"The definition of molarity is that M,"},{"Start":"00:07.770 ","End":"00:10.005","Text":"the molarity, is equal to n,"},{"Start":"00:10.005 ","End":"00:12.248","Text":"the number of moles of solute,"},{"Start":"00:12.248 ","End":"00:16.575","Text":"divided by V, the volume of solution in liters."},{"Start":"00:16.575 ","End":"00:21.990","Text":"From that, we worked out that the number of moles is given by n,"},{"Start":"00:21.990 ","End":"00:25.365","Text":"the number of moles equal to the product of M,"},{"Start":"00:25.365 ","End":"00:29.580","Text":"the molarity, and V, the volume,"},{"Start":"00:29.580 ","End":"00:36.120","Text":"which we can write without the cross as M times V moles."},{"Start":"00:36.120 ","End":"00:41.760","Text":"In this video, we will learn how to dilute solutions from"},{"Start":"00:41.760 ","End":"00:47.510","Text":"a higher molarity to a lower molarity."},{"Start":"00:47.510 ","End":"00:53.765","Text":"The key to understanding dilution is to realize that we\u0027re adding more solvent,"},{"Start":"00:53.765 ","End":"00:59.120","Text":"for example, more water without changing the amount of the solute."},{"Start":"00:59.120 ","End":"01:02.510","Text":"In other words, the number of moles,"},{"Start":"01:02.510 ","End":"01:04.685","Text":"n, doesn\u0027t change."},{"Start":"01:04.685 ","End":"01:08.230","Text":"Let\u0027s write the equation that we had before."},{"Start":"01:08.230 ","End":"01:12.540","Text":"Before we try to dilute the solution,"},{"Start":"01:12.540 ","End":"01:15.970","Text":"and we\u0027ll use 1 to indicate before."},{"Start":"01:15.970 ","End":"01:22.450","Text":"N_1=M_1V_1. Now, after we\u0027ve diluted the solution,"},{"Start":"01:22.450 ","End":"01:28.520","Text":"we can write N_2=M_2V_2, that\u0027s afterwards."},{"Start":"01:28.520 ","End":"01:33.010","Text":"Now, if the number of moles doesn\u0027t change, then n_1,"},{"Start":"01:33.010 ","End":"01:34.735","Text":"the number of moles before,"},{"Start":"01:34.735 ","End":"01:37.030","Text":"is exactly equal to n_2,"},{"Start":"01:37.030 ","End":"01:38.875","Text":"the number of moles after."},{"Start":"01:38.875 ","End":"01:40.900","Text":"We conclude from that,"},{"Start":"01:40.900 ","End":"01:47.840","Text":"that M_1V_1 is equal to M_2V_2."},{"Start":"01:47.910 ","End":"01:52.479","Text":"This is the equation we\u0027re going to use for dilution."},{"Start":"01:52.479 ","End":"01:54.295","Text":"Let\u0027s take an example."},{"Start":"01:54.295 ","End":"01:59.800","Text":"Supposing we have available a 1-molar solution of hydrochloric acid,"},{"Start":"01:59.800 ","End":"02:04.070","Text":"that\u0027s HCl in water, HCl aqueous."},{"Start":"02:04.070 ","End":"02:08.150","Text":"How much of the solution do we need to dilute to produce"},{"Start":"02:08.150 ","End":"02:13.670","Text":"250 milliliters of 0.1 molar HCl."},{"Start":"02:13.670 ","End":"02:19.630","Text":"Now, the way to tackle this problem is to write out what M_1 is,"},{"Start":"02:19.630 ","End":"02:23.350","Text":"what V_1 is, what M_2 is, what V_2 is."},{"Start":"02:23.350 ","End":"02:28.245","Text":"We\u0027ll see that we know 3 of them and we have to find the fourth one."},{"Start":"02:28.245 ","End":"02:31.175","Text":"The question tells us that M_1,"},{"Start":"02:31.175 ","End":"02:32.630","Text":"the molarity at the beginning,"},{"Start":"02:32.630 ","End":"02:34.430","Text":"was 1 molar,"},{"Start":"02:34.430 ","End":"02:38.930","Text":"1 M. After dilution, it was M_2."},{"Start":"02:38.930 ","End":"02:43.070","Text":"M_2 is equal to 0.1 M. We\u0027re diluting"},{"Start":"02:43.070 ","End":"02:47.435","Text":"from a concentrated solution of 1 M to a more dilute solution of"},{"Start":"02:47.435 ","End":"02:57.260","Text":"0.1 M. It also tells us that the volume after we\u0027ve diluted it is 250 milliliters."},{"Start":"02:57.260 ","End":"03:03.860","Text":"We want to prepare 250 milliliters of diluted hydrochloric acid."},{"Start":"03:03.860 ","End":"03:10.775","Text":"Now we can write that V_2 as 0.25 liters because remember,"},{"Start":"03:10.775 ","End":"03:14.600","Text":"in all of this, the volume was in liters."},{"Start":"03:14.600 ","End":"03:17.915","Text":"What remains for us to find out is,"},{"Start":"03:17.915 ","End":"03:19.535","Text":"what is the volume of V_1,"},{"Start":"03:19.535 ","End":"03:22.195","Text":"the volume before dilution."},{"Start":"03:22.195 ","End":"03:25.770","Text":"Now, from M_1V_1=M_2V_2,"},{"Start":"03:25.840 ","End":"03:35.350","Text":"we can write that V_1 is equal to M_2 divided by M_1 times V_2."},{"Start":"03:35.350 ","End":"03:38.990","Text":"In other words, in order to find V_1,"},{"Start":"03:38.990 ","End":"03:42.430","Text":"we\u0027re just dividing the whole equation by M_1."},{"Start":"03:42.430 ","End":"03:48.175","Text":"We have M_2 divided by M_1 multiplied by V_2."},{"Start":"03:48.175 ","End":"03:51.709","Text":"All that remains is to insert the numbers."},{"Start":"03:51.709 ","End":"03:57.070","Text":"M_2 is 0.1 M,"},{"Start":"03:57.070 ","End":"04:00.410","Text":"M_1 is 1 M,"},{"Start":"04:00.410 ","End":"04:04.290","Text":"V_2 is 0.25 liters."},{"Start":"04:04.400 ","End":"04:11.430","Text":"Now, 0.1 divided by 1 is of course just 0.1."},{"Start":"04:11.430 ","End":"04:15.698","Text":"M cancels with the M,"},{"Start":"04:15.698 ","End":"04:21.615","Text":"and we\u0027re left with 0.1 times 0.25,"},{"Start":"04:21.615 ","End":"04:26.775","Text":"that\u0027s 0.025, and the units are just liters."},{"Start":"04:26.775 ","End":"04:30.260","Text":"We have 0.025 liters."},{"Start":"04:30.260 ","End":"04:33.920","Text":"If we want to write that in milliliters,"},{"Start":"04:33.920 ","End":"04:38.795","Text":"we can calculate that as 25 milliliters."},{"Start":"04:38.795 ","End":"04:43.640","Text":"The answer is 25 milliliters."},{"Start":"04:43.640 ","End":"04:47.285","Text":"We could have done it a slightly different way."},{"Start":"04:47.285 ","End":"04:52.685","Text":"We can note that the units of V_1 and V_2 are the same."},{"Start":"04:52.685 ","End":"04:57.245","Text":"It doesn\u0027t really matter whether we take milliliters or liters."},{"Start":"04:57.245 ","End":"04:58.805","Text":"How do we know that?"},{"Start":"04:58.805 ","End":"05:04.055","Text":"Because V_1=M_2/M_1, which is just a ratio,"},{"Start":"05:04.055 ","End":"05:09.980","Text":"has no units after we\u0027ve canceled the capital M and it\u0027s multiplied by V_2."},{"Start":"05:09.980 ","End":"05:13.495","Text":"V_1 and V_2 have the same units."},{"Start":"05:13.495 ","End":"05:15.405","Text":"We could have done it like this,"},{"Start":"05:15.405 ","End":"05:18.315","Text":"M_1V_1 is equal to M_2V_2."},{"Start":"05:18.315 ","End":"05:23.535","Text":"Again, V_1 is M_2 divided by M_1 times V_2,"},{"Start":"05:23.535 ","End":"05:27.915","Text":"0.1 for M_2, 1 for M_1."},{"Start":"05:27.915 ","End":"05:29.790","Text":"Remember the Ms cancel."},{"Start":"05:29.790 ","End":"05:33.605","Text":"If we take V_2 as 250 milliliters,"},{"Start":"05:33.605 ","End":"05:38.480","Text":"then we\u0027ll have 0.1 times 250 milliliters,"},{"Start":"05:38.480 ","End":"05:41.180","Text":"which is just 25 milliliters."},{"Start":"05:41.180 ","End":"05:44.135","Text":"We got precisely the same answer,"},{"Start":"05:44.135 ","End":"05:48.545","Text":"whether we take liters or milliliters."},{"Start":"05:48.545 ","End":"05:51.140","Text":"Now, how would you do this experimentally?"},{"Start":"05:51.140 ","End":"05:56.735","Text":"We would take 25 milliliters of the 1-molar solution"},{"Start":"05:56.735 ","End":"06:03.380","Text":"in a pipette or little dropper and place it in the bottom of a volumetric flask."},{"Start":"06:03.380 ","End":"06:06.860","Text":"Remember we have volumetric flask of all different volumes."},{"Start":"06:06.860 ","End":"06:11.845","Text":"Here, we need one that holds 250 milliliters."},{"Start":"06:11.845 ","End":"06:19.940","Text":"Then all we have to do is to add water right up to the 250 milliliter mark on the neck."},{"Start":"06:19.940 ","End":"06:25.820","Text":"Remember, we just have to add water right up to this mark,"},{"Start":"06:25.820 ","End":"06:28.070","Text":"and then we have our solution."},{"Start":"06:28.070 ","End":"06:33.480","Text":"In this video, we learned how to dilute a concentrated solution."}],"ID":17671},{"Watched":false,"Name":"Exercise 13","Duration":"1m 39s","ChapterTopicVideoID":22856,"CourseChapterTopicPlaylistID":86817,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22856.jpeg","UploadDate":"2020-12-15T05:55:43.1130000","DurationForVideoObject":"PT1M39S","Description":null,"VideoComments":[],"Subtitles":[],"ID":23677},{"Watched":false,"Name":"Exercise 14","Duration":"2m 23s","ChapterTopicVideoID":22870,"CourseChapterTopicPlaylistID":86817,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22870.jpeg","UploadDate":"2020-12-15T06:00:12.8270000","DurationForVideoObject":"PT2M23S","Description":null,"VideoComments":[],"Subtitles":[],"ID":23691},{"Watched":false,"Name":"Exercise 15","Duration":"4m 32s","ChapterTopicVideoID":22873,"CourseChapterTopicPlaylistID":86817,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22873.jpeg","UploadDate":"2020-12-15T06:00:47.8900000","DurationForVideoObject":"PT4M32S","Description":null,"VideoComments":[],"Subtitles":[],"ID":23694},{"Watched":false,"Name":"Exercise 16","Duration":"5m 11s","ChapterTopicVideoID":22874,"CourseChapterTopicPlaylistID":86817,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22874.jpeg","UploadDate":"2020-12-15T06:01:02.6600000","DurationForVideoObject":"PT5M11S","Description":null,"VideoComments":[],"Subtitles":[],"ID":23695},{"Watched":false,"Name":"Exercise 17","Duration":"3m 33s","ChapterTopicVideoID":22872,"CourseChapterTopicPlaylistID":86817,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22872.jpeg","UploadDate":"2020-12-15T06:00:35.3630000","DurationForVideoObject":"PT3M33S","Description":null,"VideoComments":[],"Subtitles":[],"ID":23693},{"Watched":false,"Name":"Exercise 18","Duration":"6m ","ChapterTopicVideoID":22875,"CourseChapterTopicPlaylistID":86817,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22875.jpeg","UploadDate":"2020-12-15T06:01:19.6770000","DurationForVideoObject":"PT6M","Description":null,"VideoComments":[],"Subtitles":[],"ID":23696},{"Watched":false,"Name":"Exercise 19","Duration":"2m 59s","ChapterTopicVideoID":22854,"CourseChapterTopicPlaylistID":86817,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22854.jpeg","UploadDate":"2020-12-15T05:55:19.5230000","DurationForVideoObject":"PT2M59S","Description":null,"VideoComments":[],"Subtitles":[],"ID":23675},{"Watched":false,"Name":"Exercise 20","Duration":"5m 19s","ChapterTopicVideoID":22855,"CourseChapterTopicPlaylistID":86817,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22855.jpeg","UploadDate":"2020-12-15T05:55:38.0730000","DurationForVideoObject":"PT5M19S","Description":null,"VideoComments":[],"Subtitles":[],"ID":23676},{"Watched":false,"Name":"Exercise 21","Duration":"8m 39s","ChapterTopicVideoID":22876,"CourseChapterTopicPlaylistID":86817,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22876.jpeg","UploadDate":"2020-12-15T06:01:44.4570000","DurationForVideoObject":"PT8M39S","Description":null,"VideoComments":[],"Subtitles":[],"ID":23697},{"Watched":false,"Name":"Reaction Yields - Theory","Duration":"3m 34s","ChapterTopicVideoID":16927,"CourseChapterTopicPlaylistID":86817,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/16927.jpeg","UploadDate":"2019-02-20T23:54:33.8030000","DurationForVideoObject":"PT3M34S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.260 ","End":"00:04.260","Text":"In previous videos, we use the combustion of"},{"Start":"00:04.260 ","End":"00:07.545","Text":"glucose to explain several portrait concepts."},{"Start":"00:07.545 ","End":"00:09.855","Text":"Here\u0027s the combustion of glucose."},{"Start":"00:09.855 ","End":"00:13.620","Text":"We saw that 1 mole of glucose reacts with"},{"Start":"00:13.620 ","End":"00:17.925","Text":"6 moles of oxygen to give 6 moles of carbon dioxide,"},{"Start":"00:17.925 ","End":"00:19.890","Text":"and 6 moles of water."},{"Start":"00:19.890 ","End":"00:22.710","Text":"We used it for several purposes."},{"Start":"00:22.710 ","End":"00:26.490","Text":"We used it to balance chemical reactions."},{"Start":"00:26.490 ","End":"00:31.875","Text":"We used it to determine the formula of an unknown organic compound."},{"Start":"00:31.875 ","End":"00:35.825","Text":"We used it to talk about stoichiometry."},{"Start":"00:35.825 ","End":"00:40.025","Text":"We used it in our discussion of limiting reactants."},{"Start":"00:40.025 ","End":"00:46.790","Text":"In this video, we\u0027re going to use the same reaction in a discussion of a reaction yield,"},{"Start":"00:46.790 ","End":"00:50.345","Text":"how much of the product do we get in a reaction?"},{"Start":"00:50.345 ","End":"00:53.405","Text":"Let\u0027s begin with the theoretical yield."},{"Start":"00:53.405 ","End":"00:56.330","Text":"The theoretical yield is the amount of the product"},{"Start":"00:56.330 ","End":"00:59.585","Text":"predicted by the stoichiometry of the reaction."},{"Start":"00:59.585 ","End":"01:05.750","Text":"That\u0027s the maximum amount we could possibly obtain if everything goes as it should."},{"Start":"01:05.750 ","End":"01:07.460","Text":"Here\u0027s an example."},{"Start":"01:07.460 ","End":"01:10.280","Text":"Supposing we burn 1 mole of glucose,"},{"Start":"01:10.280 ","End":"01:12.005","Text":"just like in the equation."},{"Start":"01:12.005 ","End":"01:13.790","Text":"According to the equation,"},{"Start":"01:13.790 ","End":"01:17.945","Text":"we should obtain exactly 6 moles of carbon dioxide."},{"Start":"01:17.945 ","End":"01:22.204","Text":"This is called the theoretical yield of carbon dioxide."},{"Start":"01:22.204 ","End":"01:25.895","Text":"It isn\u0027t possible to get more carbon dioxide."},{"Start":"01:25.895 ","End":"01:28.966","Text":"But in fact, often things go wrong,"},{"Start":"01:28.966 ","End":"01:31.475","Text":"and we don\u0027t get the full amount."},{"Start":"01:31.475 ","End":"01:34.820","Text":"We get what we call the actual yield."},{"Start":"01:34.820 ","End":"01:40.070","Text":"The actual yield is the amount of product actually obtained in an experiment."},{"Start":"01:40.070 ","End":"01:44.075","Text":"There can be many reasons why we don\u0027t obtain the full amount."},{"Start":"01:44.075 ","End":"01:45.635","Text":"Perhaps there was a leak,"},{"Start":"01:45.635 ","End":"01:47.900","Text":"perhaps we drop something on the floor."},{"Start":"01:47.900 ","End":"01:52.744","Text":"Perhaps there were other reactions that took away some of our product."},{"Start":"01:52.744 ","End":"01:54.560","Text":"Let\u0027s take an example."},{"Start":"01:54.560 ","End":"01:56.220","Text":"Again, glucose."},{"Start":"01:56.220 ","End":"01:59.705","Text":"Suppose we burn 1 mole of glucose and obtain"},{"Start":"01:59.705 ","End":"02:04.138","Text":"only 5 moles of carbon dioxide because there was a leak,"},{"Start":"02:04.138 ","End":"02:05.855","Text":"and some leaked out."},{"Start":"02:05.855 ","End":"02:08.390","Text":"Recall this, the actual yield,"},{"Start":"02:08.390 ","End":"02:10.880","Text":"the actual amount we obtained."},{"Start":"02:10.880 ","End":"02:14.000","Text":"Now we define the percentage yield,"},{"Start":"02:14.000 ","End":"02:19.085","Text":"the actual percentage that we get out in the reaction."},{"Start":"02:19.085 ","End":"02:20.855","Text":"Here\u0027s the definition."},{"Start":"02:20.855 ","End":"02:24.485","Text":"The percentage yield is the actual yield"},{"Start":"02:24.485 ","End":"02:29.540","Text":"divided by the theoretical yield multiplied by a 100 percent."},{"Start":"02:29.540 ","End":"02:35.150","Text":"If we got all that we should have if the actual yield was equal to the theoretical yield,"},{"Start":"02:35.150 ","End":"02:37.805","Text":"this would give us a 100 percent."},{"Start":"02:37.805 ","End":"02:39.995","Text":"Let\u0027s take the example."},{"Start":"02:39.995 ","End":"02:43.520","Text":"Where the actual yield of carbon dioxide is 5 moles,"},{"Start":"02:43.520 ","End":"02:46.430","Text":"the percentage yield is 5 moles."},{"Start":"02:46.430 ","End":"02:48.020","Text":"That\u0027s the actual yield,"},{"Start":"02:48.020 ","End":"02:50.420","Text":"divided by 6 moles,"},{"Start":"02:50.420 ","End":"02:53.125","Text":"which is the theoretical yield."},{"Start":"02:53.125 ","End":"02:57.686","Text":"And I\u0027ve missed out the moles because they cancel."},{"Start":"02:57.686 ","End":"02:59.225","Text":"Multiplied by a 100 percent."},{"Start":"02:59.225 ","End":"03:04.710","Text":"We work this out, it\u0027s only 83.3 percent."},{"Start":"03:06.700 ","End":"03:13.295","Text":"The percentage yield is significantly smaller than a 100 percent."},{"Start":"03:13.295 ","End":"03:15.395","Text":"This is usually the case."},{"Start":"03:15.395 ","End":"03:19.070","Text":"It\u0027s very rare to get a 100 percent."},{"Start":"03:19.070 ","End":"03:24.065","Text":"In this video, we talked about the theoretical yield,"},{"Start":"03:24.065 ","End":"03:27.485","Text":"the actual yield, and the percentage yield."},{"Start":"03:27.485 ","End":"03:28.850","Text":"In the next video,"},{"Start":"03:28.850 ","End":"03:33.750","Text":"I\u0027ll give a more detailed example of the concept."}],"ID":17672},{"Watched":false,"Name":"Reaction Yields - Example","Duration":"7m 37s","ChapterTopicVideoID":16928,"CourseChapterTopicPlaylistID":86817,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/16928.jpeg","UploadDate":"2019-02-20T23:55:19.8600000","DurationForVideoObject":"PT7M37S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.530","Text":"In the previous video,"},{"Start":"00:01.530 ","End":"00:04.305","Text":"we discussed the theory of reaction yields."},{"Start":"00:04.305 ","End":"00:07.790","Text":"In this video, I will give an example of how to calculate them."},{"Start":"00:07.790 ","End":"00:09.300","Text":"Here\u0027s our example."},{"Start":"00:09.300 ","End":"00:16.275","Text":"6 grams of solid sulfur reacts with 12 grams of oxygen gas to produce SO_3,"},{"Start":"00:16.275 ","End":"00:19.320","Text":"sulfur trioxide, which is also a gas."},{"Start":"00:19.320 ","End":"00:22.860","Text":"Calculate the theoretical yield of SO_3."},{"Start":"00:22.860 ","End":"00:28.485","Text":"If 12.20 grams is obtained in an experiment,"},{"Start":"00:28.485 ","End":"00:30.810","Text":"calculate the percentage yield."},{"Start":"00:30.810 ","End":"00:32.160","Text":"Before we go on,"},{"Start":"00:32.160 ","End":"00:37.095","Text":"let\u0027s remember that what\u0027s obtained in the experiment is the actual yield."},{"Start":"00:37.095 ","End":"00:40.545","Text":"This number is the actual yield."},{"Start":"00:40.545 ","End":"00:45.710","Text":"The very first step is to write and balance the chemical equation."},{"Start":"00:45.710 ","End":"00:49.955","Text":"S_8, that\u0027s sulfur in its elemental state,"},{"Start":"00:49.955 ","End":"00:54.020","Text":"plus O_2 gives SO_3."},{"Start":"00:54.020 ","End":"00:56.990","Text":"Now, let\u0027s look at the equation."},{"Start":"00:56.990 ","End":"01:02.300","Text":"We see we have 8 sulfurs here and only 1 on the right-hand side,"},{"Start":"01:02.300 ","End":"01:06.490","Text":"so we need to multiply SO_3 by 8."},{"Start":"01:06.490 ","End":"01:09.165","Text":"Let\u0027s look at the oxygen."},{"Start":"01:09.165 ","End":"01:15.470","Text":"We have 8 times 3 oxygens on the right and only 2 on the left,"},{"Start":"01:15.470 ","End":"01:20.790","Text":"so we need 12 times 2 to get to 24."},{"Start":"01:20.790 ","End":"01:24.690","Text":"We have now 24 on the left and 24 on the right,"},{"Start":"01:24.690 ","End":"01:27.390","Text":"so we need to write here 12."},{"Start":"01:27.390 ","End":"01:29.565","Text":"Here, it\u0027s written better."},{"Start":"01:29.565 ","End":"01:33.380","Text":"S_8 solid plus 12 oxygen gas,"},{"Start":"01:33.380 ","End":"01:36.185","Text":"giving us 8 SO_3 gas."},{"Start":"01:36.185 ","End":"01:39.935","Text":"Now, we can write out our scheme for solving the problem."},{"Start":"01:39.935 ","End":"01:42.080","Text":"We\u0027re given the mass of reactants,"},{"Start":"01:42.080 ","End":"01:44.270","Text":"we can calculate the moles of reactants."},{"Start":"01:44.270 ","End":"01:46.265","Text":"That\u0027s step 2."},{"Start":"01:46.265 ","End":"01:53.375","Text":"Then we have to decide whether there\u0027s a limiting reactant, that step 2a."},{"Start":"01:53.375 ","End":"01:58.625","Text":"Then we\u0027d go from the moles of reactants to the moles of products in here,"},{"Start":"01:58.625 ","End":"02:01.595","Text":"counter the limiting reactant if there is one."},{"Start":"02:01.595 ","End":"02:03.290","Text":"That\u0027s step 3."},{"Start":"02:03.290 ","End":"02:07.238","Text":"For the moles of the product to the mass of product, that\u0027s step 4."},{"Start":"02:07.238 ","End":"02:13.505","Text":"The mass of the product given us here will be the theoretical yield."},{"Start":"02:13.505 ","End":"02:16.535","Text":"Then finally, step 5,"},{"Start":"02:16.535 ","End":"02:21.360","Text":"regards to the percentage yield. Let\u0027s do it."},{"Start":"02:22.250 ","End":"02:27.365","Text":"Step 2, we have to calculate the moles of reactants."},{"Start":"02:27.365 ","End":"02:28.565","Text":"We\u0027re given the masses,"},{"Start":"02:28.565 ","End":"02:31.265","Text":"we need to convert them to moles."},{"Start":"02:31.265 ","End":"02:34.040","Text":"We need to know the molar masses."},{"Start":"02:34.040 ","End":"02:38.479","Text":"The molar mass of S_8 is 256.5 grams per mole,"},{"Start":"02:38.479 ","End":"02:43.490","Text":"and the molar mass of O_2 is 32.00 grams per mole."},{"Start":"02:43.490 ","End":"02:46.190","Text":"Now, we can calculate the number of moles."},{"Start":"02:46.190 ","End":"02:52.840","Text":"The number of moles of acid is the mass of acid divided by the molar mass,"},{"Start":"02:52.840 ","End":"02:58.988","Text":"so 6.00 grams divided by 256.5 grams per mole,"},{"Start":"02:58.988 ","End":"03:03.685","Text":"and that gives us 0.02339 moles."},{"Start":"03:03.685 ","End":"03:06.640","Text":"The number of moles of oxygen is 12 grams,"},{"Start":"03:06.640 ","End":"03:08.935","Text":"that\u0027s how much the mass of oxygen,"},{"Start":"03:08.935 ","End":"03:10.630","Text":"divided by the molar mass,"},{"Start":"03:10.630 ","End":"03:13.285","Text":"32.00 grams per mole."},{"Start":"03:13.285 ","End":"03:15.415","Text":"Again, the grams goes,"},{"Start":"03:15.415 ","End":"03:17.365","Text":"grams cancels with grams,"},{"Start":"03:17.365 ","End":"03:21.220","Text":"we\u0027re left with 1 over moles minus 1 which is moles."},{"Start":"03:21.220 ","End":"03:28.360","Text":"So 12 divided by 32 gives us 0.3750 moles."},{"Start":"03:28.360 ","End":"03:34.190","Text":"We have to look and see whether we have a limiting reactant or a limiting reagent,"},{"Start":"03:34.190 ","End":"03:35.750","Text":"some people call it."},{"Start":"03:35.750 ","End":"03:39.710","Text":"Supposing we want to use all the oxygen,"},{"Start":"03:39.710 ","End":"03:43.865","Text":"we want to use 0.3750 moles of oxygen,"},{"Start":"03:43.865 ","End":"03:47.975","Text":"how many moles of sulfur would we require?"},{"Start":"03:47.975 ","End":"03:52.565","Text":"The moles of sulfur is equal to the moles of oxygen"},{"Start":"03:52.565 ","End":"03:59.600","Text":"multiplied by the stoichiometric ratio, or stoichiometric factor."},{"Start":"03:59.600 ","End":"04:06.455","Text":"1 mole of S_8 requires 12 moles of oxygen to react."},{"Start":"04:06.455 ","End":"04:08.300","Text":"Now we can calculate it."},{"Start":"04:08.300 ","End":"04:14.540","Text":"We can substitute the value of the number of moles of oxygen, 0.3750."},{"Start":"04:14.540 ","End":"04:19.615","Text":"Here\u0027s 0.3750 divided by 12,"},{"Start":"04:19.615 ","End":"04:24.830","Text":"and the units are moles of sulfur because moles of oxygen goes with moles of the oxygen,"},{"Start":"04:24.830 ","End":"04:26.870","Text":"we\u0027re left with moles of sulfur."},{"Start":"04:26.870 ","End":"04:31.400","Text":"We have 0.300750 divided by 12 moles of sulfur,"},{"Start":"04:31.400 ","End":"04:37.870","Text":"and if we divide that we get 0.03125 moles of sulfur."},{"Start":"04:37.870 ","End":"04:42.140","Text":"This tells us that in order to use up all the oxygen,"},{"Start":"04:42.140 ","End":"04:46.789","Text":"we require 0.03125 moles of sulfur."},{"Start":"04:46.789 ","End":"04:50.135","Text":"In the step before the second step,"},{"Start":"04:50.135 ","End":"04:51.560","Text":"we saw this,"},{"Start":"04:51.560 ","End":"04:55.790","Text":"in fact we only have 0.02339 moles"},{"Start":"04:55.790 ","End":"05:00.920","Text":"so we don\u0027t have enough sulfur for all the oxygen to react."},{"Start":"05:00.920 ","End":"05:03.590","Text":"Sulfur is the limiting reactant;"},{"Start":"05:03.590 ","End":"05:07.565","Text":"it will all be used up and some oxygen will be left."},{"Start":"05:07.565 ","End":"05:09.695","Text":"Oxygen is in excess."},{"Start":"05:09.695 ","End":"05:13.460","Text":"Step 3 is to calculate the moles of SO_3."},{"Start":"05:13.460 ","End":"05:21.170","Text":"The moles of SO_3 is number of moles of sulfur multiplied by the stoichiometric factor."},{"Start":"05:21.170 ","End":"05:23.195","Text":"For every mole of S_8,"},{"Start":"05:23.195 ","End":"05:25.895","Text":"we get 8 moles of SO_3."},{"Start":"05:25.895 ","End":"05:27.470","Text":"So it\u0027s 8 to 1."},{"Start":"05:27.470 ","End":"05:30.544","Text":"Now, we can substitute the numbers."},{"Start":"05:30.544 ","End":"05:37.460","Text":"The number of moles of S_8 that we have in practice is 0.02339 moles,"},{"Start":"05:37.460 ","End":"05:43.550","Text":"and that\u0027s multiplied by 8, giving us 0.1871."},{"Start":"05:43.550 ","End":"05:45.245","Text":"What about the units?"},{"Start":"05:45.245 ","End":"05:48.425","Text":"Moles of sulfur goes with moles of sulfur,"},{"Start":"05:48.425 ","End":"05:51.200","Text":"moles of sulfur here goes from moles of sulfur,"},{"Start":"05:51.200 ","End":"05:55.330","Text":"so we\u0027re left with just moles of SO_3."},{"Start":"05:55.330 ","End":"06:01.860","Text":"Our answer is 0.1871 moles of SO_3."},{"Start":"06:01.860 ","End":"06:06.750","Text":"We need to go from moles SO_3 to the mass of SO_3."},{"Start":"06:06.750 ","End":"06:09.935","Text":"To do that, we need the molar mass."},{"Start":"06:09.935 ","End":"06:14.700","Text":"The molar mass of SO_3 is 80.07 grams per mole,"},{"Start":"06:14.700 ","End":"06:17.510","Text":"and the mass, remember,"},{"Start":"06:17.510 ","End":"06:20.255","Text":"is the number of moles times the molar mass."},{"Start":"06:20.255 ","End":"06:25.355","Text":"That\u0027s 0.1871 moles as we calculated before,"},{"Start":"06:25.355 ","End":"06:28.760","Text":"times 80.07 grams per mole."},{"Start":"06:28.760 ","End":"06:30.695","Text":"Multiply the 2 numbers,"},{"Start":"06:30.695 ","End":"06:38.700","Text":"0.1871 times 80.07, we get 14.98 grams. Why grams?"},{"Start":"06:38.700 ","End":"06:42.405","Text":"Because moles times moles^minus 1 is just 1,"},{"Start":"06:42.405 ","End":"06:44.625","Text":"and we\u0027re left just with grams."},{"Start":"06:44.625 ","End":"06:48.605","Text":"Here\u0027s our answer, 14.98 grams."},{"Start":"06:48.605 ","End":"06:51.925","Text":"That is the theoretical yield."},{"Start":"06:51.925 ","End":"06:56.965","Text":"All that remains now is to calculate the percentage yield."},{"Start":"06:56.965 ","End":"06:59.840","Text":"The percentage yield is the actual yield,"},{"Start":"06:59.840 ","End":"07:03.545","Text":"that\u0027s what\u0027s given to us in the question, 12.20 grams,"},{"Start":"07:03.545 ","End":"07:08.540","Text":"divided by the theoretical yield, 14.98 grams,"},{"Start":"07:08.540 ","End":"07:11.620","Text":"which we just calculated in the step before,"},{"Start":"07:11.620 ","End":"07:13.770","Text":"times 100 percent,"},{"Start":"07:13.770 ","End":"07:16.935","Text":"and that\u0027s 81.44 percent."},{"Start":"07:16.935 ","End":"07:22.000","Text":"Here\u0027s our percentage yield, 81.44 percent."},{"Start":"07:22.000 ","End":"07:27.230","Text":"In this video, we solved an example based on the reaction yield."},{"Start":"07:27.230 ","End":"07:32.645","Text":"The solution involved many concepts that we\u0027ve learned in previous videos."},{"Start":"07:32.645 ","End":"07:36.780","Text":"It\u0027s really quite a good revision of those concepts."}],"ID":17673},{"Watched":false,"Name":"Exercise 22","Duration":"3m 15s","ChapterTopicVideoID":22877,"CourseChapterTopicPlaylistID":86817,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22877.jpeg","UploadDate":"2020-12-15T06:01:53.7070000","DurationForVideoObject":"PT3M15S","Description":null,"VideoComments":[],"Subtitles":[],"ID":23698}],"Thumbnail":null,"ID":86817},{"Name":"Reactions","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Consecutive Reactions","Duration":"6m 15s","ChapterTopicVideoID":16918,"CourseChapterTopicPlaylistID":86818,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/16918.jpeg","UploadDate":"2019-02-20T23:48:52.2300000","DurationForVideoObject":"PT6M15S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.755","Text":"In this video, we will discuss consecutive reactions."},{"Start":"00:04.755 ","End":"00:10.650","Text":"Consecutive reactions are of the general type A, some reactants,"},{"Start":"00:10.650 ","End":"00:14.295","Text":"going to B some products,"},{"Start":"00:14.295 ","End":"00:19.635","Text":"and then B, being the reactants going to C some products."},{"Start":"00:19.635 ","End":"00:26.130","Text":"B is called an intermediate since it is formed in the first reaction, A to B,"},{"Start":"00:26.130 ","End":"00:28.740","Text":"and then reacts in the second reaction,"},{"Start":"00:28.740 ","End":"00:34.515","Text":"B to C. We can add the 2 equations together to get the overall reaction."},{"Start":"00:34.515 ","End":"00:36.240","Text":"Here are 2 equations,"},{"Start":"00:36.240 ","End":"00:42.950","Text":"A to B and B to C. You see that B appears on the right-hand side in the first equation,"},{"Start":"00:42.950 ","End":"00:44.330","Text":"and on the left-hand side,"},{"Start":"00:44.330 ","End":"00:45.800","Text":"in the second equation."},{"Start":"00:45.800 ","End":"00:47.990","Text":"We can cancel them."},{"Start":"00:47.990 ","End":"00:50.585","Text":"Now we can add up the 2 equations,"},{"Start":"00:50.585 ","End":"00:55.085","Text":"A on the left-hand side and C on the right-hand side."},{"Start":"00:55.085 ","End":"00:57.140","Text":"If that isn\u0027t clear,"},{"Start":"00:57.140 ","End":"00:59.795","Text":"we can do it without canceling the B,"},{"Start":"00:59.795 ","End":"01:04.890","Text":"A plus B going to B plus"},{"Start":"01:04.890 ","End":"01:10.220","Text":"C. Now you can see that B appears in both sides of the equation,"},{"Start":"01:10.220 ","End":"01:14.105","Text":"so can be canceled, leading to A,"},{"Start":"01:14.105 ","End":"01:23.335","Text":"that gives us A going to C. Let\u0027s take an example,"},{"Start":"01:23.335 ","End":"01:28.895","Text":"write the overall reaction for the following balanced consecutive reactions."},{"Start":"01:28.895 ","End":"01:34.210","Text":"CH_4 gas, that\u0027s methane gas plus chlorine gas,"},{"Start":"01:34.210 ","End":"01:36.925","Text":"giving us methyl chloride gas,"},{"Start":"01:36.925 ","End":"01:39.310","Text":"and hydrogen chloride gas."},{"Start":"01:39.310 ","End":"01:46.120","Text":"Then CH_3Cl, that\u0027s methyl chloride reacts with chlorine"},{"Start":"01:46.120 ","End":"01:53.495","Text":"again to give us CH_2CL_2 gas plus HCL gas."},{"Start":"01:53.495 ","End":"02:00.625","Text":"We see that CH_3Cl is formed the first one and reacts in the second reaction."},{"Start":"02:00.625 ","End":"02:09.315","Text":"The second part of the question is if 10 grams of CH_4 reacts with excess CO_2,"},{"Start":"02:09.315 ","End":"02:13.060","Text":"how much CH_2CL_2 will be obtained?"},{"Start":"02:13.060 ","End":"02:16.520","Text":"The first step is to get the overall reaction."},{"Start":"02:16.520 ","End":"02:20.210","Text":"We do that by adding the 2 equations together."},{"Start":"02:20.210 ","End":"02:23.285","Text":"Here are 2 equations as we had before."},{"Start":"02:23.285 ","End":"02:25.445","Text":"Now we can add them together."},{"Start":"02:25.445 ","End":"02:30.175","Text":"We see that CH_3Cl appears on the right here,"},{"Start":"02:30.175 ","End":"02:32.325","Text":"and on the left here."},{"Start":"02:32.325 ","End":"02:33.770","Text":"It appears on the right,"},{"Start":"02:33.770 ","End":"02:35.570","Text":"it\u0027s formed in the first reaction,"},{"Start":"02:35.570 ","End":"02:37.715","Text":"and reacts in the second reaction."},{"Start":"02:37.715 ","End":"02:39.800","Text":"We can cancel it out."},{"Start":"02:39.800 ","End":"02:43.580","Text":"CH_3CL is an intermediate."},{"Start":"02:43.580 ","End":"02:47.000","Text":"Let\u0027s add the 2 to get our overall reaction."},{"Start":"02:47.000 ","End":"02:48.890","Text":"Here we have CH_4,"},{"Start":"02:48.890 ","End":"02:51.060","Text":"CH_4, CL_2,"},{"Start":"02:51.060 ","End":"02:54.075","Text":"and another CL_2 gives us 2 CL_2."},{"Start":"02:54.075 ","End":"02:58.620","Text":"CH_2, CL_2 here it appears to the right hand side."},{"Start":"02:58.620 ","End":"03:04.085","Text":"HCL in the first reaction and HCL in the second reaction gives us 2 HCL."},{"Start":"03:04.085 ","End":"03:10.220","Text":"This reaction is balanced because we added up 2 balanced reactions."},{"Start":"03:10.220 ","End":"03:13.010","Text":"This is the overall reaction."},{"Start":"03:13.010 ","End":"03:14.555","Text":"Now we can write,"},{"Start":"03:14.555 ","End":"03:17.690","Text":"the scheme for answering the question."},{"Start":"03:17.690 ","End":"03:22.500","Text":"The mass of methane goes to moles of methane."},{"Start":"03:22.500 ","End":"03:26.090","Text":"That\u0027s Step 2."},{"Start":"03:26.090 ","End":"03:29.600","Text":"Moles of methane to moles of CH_2CL_2,"},{"Start":"03:29.600 ","End":"03:31.450","Text":"that\u0027s the third step."},{"Start":"03:31.450 ","End":"03:37.320","Text":"The final step, moles of CH_2CL_2 to mass of CH_2CL_2,"},{"Start":"03:37.320 ","End":"03:39.225","Text":"that\u0027s the fourth step."},{"Start":"03:39.225 ","End":"03:43.680","Text":"Step 2 is to turn the mass of CH_4,"},{"Start":"03:43.680 ","End":"03:45.495","Text":"to moles of CH_4."},{"Start":"03:45.495 ","End":"03:48.610","Text":"This is something we\u0027ve done several times."},{"Start":"03:48.610 ","End":"03:51.185","Text":"It should be familiar to you by now."},{"Start":"03:51.185 ","End":"03:55.240","Text":"The first thing we need to know is the molar mass of CH_4."},{"Start":"03:55.240 ","End":"03:58.280","Text":"That\u0027s 16.0 grams per mole."},{"Start":"03:58.280 ","End":"04:00.710","Text":"From that, we can get the number of moles."},{"Start":"04:00.710 ","End":"04:02.630","Text":"The number of moles, if you remember,"},{"Start":"04:02.630 ","End":"04:08.665","Text":"n is equal to the mass divided by the molecular weight or the molar mass."},{"Start":"04:08.665 ","End":"04:12.350","Text":"Here we have it, n_CH_4,"},{"Start":"04:12.350 ","End":"04:19.655","Text":"the number of moles is equal to the mass 10.0 Grams divided by the molar mass,"},{"Start":"04:19.655 ","End":"04:22.055","Text":"16.0 grams per mole."},{"Start":"04:22.055 ","End":"04:25.490","Text":"That gives us 0.625 moles."},{"Start":"04:25.490 ","End":"04:29.280","Text":"Steps 3, the number of moles of CH_4,"},{"Start":"04:29.280 ","End":"04:31.810","Text":"to moles of CH_CL_2."},{"Start":"04:31.810 ","End":"04:34.580","Text":"This is the central part of the problem."},{"Start":"04:34.580 ","End":"04:36.845","Text":"Now if we look at the equation we had,"},{"Start":"04:36.845 ","End":"04:40.880","Text":"we see that for every mole of CH_4,"},{"Start":"04:40.880 ","End":"04:44.405","Text":"we get 1 mole of CH_2CL_2."},{"Start":"04:44.405 ","End":"04:51.580","Text":"The number of moles of CH_2CL_2 will be exactly equal to the number of moles of methane."},{"Start":"04:51.580 ","End":"04:56.390","Text":"N_CH_2CL_2 is the same as the number of moles of methane,"},{"Start":"04:56.390 ","End":"05:02.075","Text":"is 0.625 moles, exactly what we calculated before."},{"Start":"05:02.075 ","End":"05:03.755","Text":"Now in Step 4,"},{"Start":"05:03.755 ","End":"05:11.520","Text":"we calculate the mass of CH_2CL_2 from the number of moles of CH_2CL_2."},{"Start":"05:11.520 ","End":"05:15.655","Text":"Once again, we need the molar mass of CH_2CL_2,"},{"Start":"05:15.655 ","End":"05:19.165","Text":"and that\u0027s 84.9 grams per mole."},{"Start":"05:19.165 ","End":"05:21.410","Text":"We can work out the mass."},{"Start":"05:21.410 ","End":"05:26.000","Text":"We know that the mass is equal to the number of"},{"Start":"05:26.000 ","End":"05:33.100","Text":"moles times the molar mass, that\u0027s Mw."},{"Start":"05:33.100 ","End":"05:35.810","Text":"Here we have the number of moles,"},{"Start":"05:35.810 ","End":"05:39.365","Text":"0.625 moles, and the molar mass,"},{"Start":"05:39.365 ","End":"05:41.930","Text":"84.9 grams per mole."},{"Start":"05:41.930 ","End":"05:47.390","Text":"We can multiply the 2 numbers and get 53.1 and as for the units,"},{"Start":"05:47.390 ","End":"05:51.830","Text":"mole cancels with mole minus 1 to give 1."},{"Start":"05:51.830 ","End":"05:56.529","Text":"We have just the grams remaining, grams, grams."},{"Start":"05:56.529 ","End":"05:59.380","Text":"That\u0027s our final answer."},{"Start":"05:59.380 ","End":"06:04.990","Text":"The mass of CH_2CL_2 is 53.1 grams."},{"Start":"06:04.990 ","End":"06:09.050","Text":"In this video we learned about consecutive reactions,"},{"Start":"06:09.050 ","End":"06:10.730","Text":"in the next video,"},{"Start":"06:10.730 ","End":"06:15.150","Text":"we\u0027ll learn about simultaneous reactions."}],"ID":17665},{"Watched":false,"Name":"Exercise 1 - Part a","Duration":"6m 48s","ChapterTopicVideoID":22878,"CourseChapterTopicPlaylistID":86818,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22878.jpeg","UploadDate":"2020-12-15T06:11:07.7870000","DurationForVideoObject":"PT6M48S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.385","Text":"Hi. We are going to solve the following exercise."},{"Start":"00:02.385 ","End":"00:05.310","Text":"How many grams of hydrochloric acid are consumed in"},{"Start":"00:05.310 ","End":"00:08.520","Text":"the reaction of 415 grams of a mixture"},{"Start":"00:08.520 ","End":"00:16.035","Text":"containing 33.5 percent magnesium carbonate and 66.5 percent magnesium hydroxide by mass?"},{"Start":"00:16.035 ","End":"00:21.240","Text":"Magnesium hydroxide plus 2 hydrogen chloride gives us magnesium chloride plus 2 water."},{"Start":"00:21.240 ","End":"00:24.000","Text":"Magnesium carbonate plus 2 hydrogen chloride gives us"},{"Start":"00:24.000 ","End":"00:27.240","Text":"magnesium chloride plus water plus carbon dioxide."},{"Start":"00:27.240 ","End":"00:28.590","Text":"If we look at the equation,"},{"Start":"00:28.590 ","End":"00:31.260","Text":"we see that the hydrochloric acid is present in"},{"Start":"00:31.260 ","End":"00:33.360","Text":"both equations and to calculate"},{"Start":"00:33.360 ","End":"00:35.580","Text":"the mass of the hydrogen chloride, which is what we need to do."},{"Start":"00:35.580 ","End":"00:37.755","Text":"We need to know the grams of hydrogen chloride."},{"Start":"00:37.755 ","End":"00:39.590","Text":"We\u0027re going to first calculate the number of moles of"},{"Start":"00:39.590 ","End":"00:42.530","Text":"magnesium hydroxide and magnesium carbonate."},{"Start":"00:42.530 ","End":"00:45.770","Text":"Then we will calculate the number of moles of hydrogen chloride,"},{"Start":"00:45.770 ","End":"00:49.039","Text":"and then we\u0027ll convert the number of moles of hydrogen chloride to mass."},{"Start":"00:49.039 ","End":"00:53.755","Text":"We\u0027ll begin by calculating the moles of magnesium hydroxide and magnesium carbonate."},{"Start":"00:53.755 ","End":"00:57.005","Text":"From the question, we know that the mass of the mixture"},{"Start":"00:57.005 ","End":"01:00.470","Text":"of the magnesium hydroxide and magnesium carbonate is 415 grams,"},{"Start":"01:00.470 ","End":"01:05.690","Text":"so the mass of magnesium carbonate plus the mass of"},{"Start":"01:05.690 ","End":"01:11.615","Text":"the magnesium hydroxide equals 415 grams."},{"Start":"01:11.615 ","End":"01:15.320","Text":"We also know their percent inside the mixture."},{"Start":"01:15.320 ","End":"01:17.900","Text":"To calculate the mass,"},{"Start":"01:17.900 ","End":"01:21.120","Text":"let\u0027s take the magnesium hydroxide."},{"Start":"01:21.410 ","End":"01:25.030","Text":"For example, the equals their mixture mass,"},{"Start":"01:25.030 ","End":"01:28.635","Text":"which is 415 grams times,"},{"Start":"01:28.635 ","End":"01:30.460","Text":"look at the magnesium hydroxide,"},{"Start":"01:30.460 ","End":"01:34.330","Text":"we can see it as 66.5 percent of the mixture."},{"Start":"01:34.330 ","End":"01:40.284","Text":"Meaning if we have 100 grams of mixture and 66.5 grams of the mixture,"},{"Start":"01:40.284 ","End":"01:43.550","Text":"will be magnesium hydroxide."},{"Start":"01:43.650 ","End":"01:53.240","Text":"If we have 100 grams of mixture and this is 415 grams of the mixture."},{"Start":"01:53.730 ","End":"01:57.805","Text":"The grams of the mixture cancel out,"},{"Start":"01:57.805 ","End":"02:08.900","Text":"and 415 times 66.5 divided by 100 gives us 275.98 grams of magnesium hydroxide."},{"Start":"02:10.180 ","End":"02:13.430","Text":"Now we know the mass of the magnesium hydroxide."},{"Start":"02:13.430 ","End":"02:16.620","Text":"Lets find the mass of the magnesium carbonate."},{"Start":"02:19.060 ","End":"02:22.670","Text":"All we need to do is take their mass of the whole mixture,"},{"Start":"02:22.670 ","End":"02:29.130","Text":"which is 415 grams minus the mass of the magnesium hydroxide,"},{"Start":"02:31.640 ","End":"02:33.720","Text":"which we just calculated,"},{"Start":"02:33.720 ","End":"02:41.010","Text":"so it equals 415 grams minus 275.98 grams,"},{"Start":"02:41.010 ","End":"02:49.100","Text":"and we get 139.02 grams."},{"Start":"02:49.100 ","End":"02:53.554","Text":"We have the mass of the magnesium hydroxide and the mass of the magnesium carbonates."},{"Start":"02:53.554 ","End":"02:58.340","Text":"The next step is to calculate the moles of magnesium hydroxide and magnesium carbonate."},{"Start":"02:58.340 ","End":"03:03.440","Text":"Calculated the mass of the magnesium hydroxide and magnesium carbonates and we got that"},{"Start":"03:03.440 ","End":"03:10.860","Text":"the mass of the magnesium hydroxide equals to 75.98 grams."},{"Start":"03:10.860 ","End":"03:20.280","Text":"The mass of the magnesium carbonate equals 139.02 grams."},{"Start":"03:21.110 ","End":"03:23.570","Text":"Now the next step is to calculate"},{"Start":"03:23.570 ","End":"03:27.470","Text":"the moles of the magnesium hydroxide and magnesium carbonate."},{"Start":"03:27.470 ","End":"03:29.870","Text":"For this purpose, we\u0027re going to use the equation,"},{"Start":"03:29.870 ","End":"03:32.060","Text":"n number of moles equals m,"},{"Start":"03:32.060 ","End":"03:35.360","Text":"which is the mass divided by the molar mass."},{"Start":"03:35.360 ","End":"03:38.000","Text":"We\u0027re going to start with the magnesium hydroxide."},{"Start":"03:38.000 ","End":"03:43.115","Text":"We know its mass because it is calculated now we need to find the molar mass."},{"Start":"03:43.115 ","End":"03:48.920","Text":"The molar mass of magnesium hydroxide equals the molar mass of magnesium,"},{"Start":"03:48.920 ","End":"03:53.014","Text":"because we have 1 magnesium plus 2 times the molar mass of oxygen."},{"Start":"03:53.014 ","End":"03:57.785","Text":"We have 2 oxygens plus 2 times the molar mass of hydrogen,"},{"Start":"03:57.785 ","End":"03:59.165","Text":"because we have 2 hydrogens."},{"Start":"03:59.165 ","End":"04:07.145","Text":"This equals 24.31 grams per mole plus 2 times 16 grams per mole,"},{"Start":"04:07.145 ","End":"04:12.110","Text":"plus 2 times 1.01 grams per mole"},{"Start":"04:12.110 ","End":"04:19.230","Text":"and it equals 58.33 grams per mole."},{"Start":"04:21.160 ","End":"04:23.750","Text":"Now we can calculate the number of moles."},{"Start":"04:23.750 ","End":"04:26.150","Text":"The number of moles of"},{"Start":"04:26.150 ","End":"04:29.600","Text":"magnesium hydroxide equals the mass"},{"Start":"04:29.600 ","End":"04:32.660","Text":"of magnesium hydroxide divided by the molar mass of magnesium hydroxide,"},{"Start":"04:32.660 ","End":"04:39.545","Text":"which equals here 275.98 grams divided by the molar mass,"},{"Start":"04:39.545 ","End":"04:44.180","Text":"which is 58.33 grams per mole."},{"Start":"04:47.270 ","End":"04:51.280","Text":"This equals 4.73 moles."},{"Start":"04:51.720 ","End":"04:54.145","Text":"Just a little reminder,"},{"Start":"04:54.145 ","End":"04:56.200","Text":"when we did this in previous videos,"},{"Start":"04:56.200 ","End":"04:58.390","Text":"when you divide by a fraction, for example,"},{"Start":"04:58.390 ","End":"05:00.520","Text":"grams divided by grams per mole,"},{"Start":"05:00.520 ","End":"05:04.250","Text":"it is the same as multiplying by the reciprocal of the fraction,"},{"Start":"05:04.250 ","End":"05:06.610","Text":"so grams times mole per grams,"},{"Start":"05:06.610 ","End":"05:09.460","Text":"grams cancel out and we\u0027re left with mole."},{"Start":"05:09.460 ","End":"05:13.480","Text":"The moles of magnesium hydroxide are 4.73."},{"Start":"05:13.480 ","End":"05:16.959","Text":"Now we\u0027re going to calculate the moles of magnesium carbonate."},{"Start":"05:16.959 ","End":"05:19.750","Text":"Again, we need the molar mass of magnesium carbonate,"},{"Start":"05:19.750 ","End":"05:25.090","Text":"so the molar mass of magnesium carbonate equals the molar mass of"},{"Start":"05:25.090 ","End":"05:33.515","Text":"magnesium plus the molar mass of the carbon plus 3 times the molar mass of oxygen."},{"Start":"05:33.515 ","End":"05:41.945","Text":"This equals 24.31 grams per mole plus 12 grams per mole,"},{"Start":"05:41.945 ","End":"05:44.359","Text":"plus 3 times 16 grams per mole,"},{"Start":"05:44.359 ","End":"05:51.920","Text":"and this equals 84.31 grams per mole."},{"Start":"05:51.920 ","End":"05:54.845","Text":"That\u0027s the molar mass of magnesium carbonate."},{"Start":"05:54.845 ","End":"05:58.325","Text":"Now we can calculate the number of moles of magnesium carbonate."},{"Start":"05:58.325 ","End":"06:03.140","Text":"N, the number of moles equals the mass divided by the molar mass."},{"Start":"06:03.140 ","End":"06:05.705","Text":"We\u0027re talking about magnesium carbonate."},{"Start":"06:05.705 ","End":"06:09.590","Text":"The mass of the magnesium carbonate is what we calculated in the previous video,"},{"Start":"06:09.590 ","End":"06:13.730","Text":"n equals 139.02 grams divided by the molar mass,"},{"Start":"06:13.730 ","End":"06:19.680","Text":"which is 84.31 grams per mole, of course."},{"Start":"06:20.260 ","End":"06:24.005","Text":"This equals 1.65 mole."},{"Start":"06:24.005 ","End":"06:25.894","Text":"Of course, in both cases,"},{"Start":"06:25.894 ","End":"06:31.453","Text":"the molar masses were taken from the periodic table of elements."},{"Start":"06:31.453 ","End":"06:35.000","Text":"Now we have the moles of the magnesium hydroxide,"},{"Start":"06:35.000 ","End":"06:39.230","Text":"which is 4.73 moles and the moles of magnesium carbonate,"},{"Start":"06:39.230 ","End":"06:41.105","Text":"which is 1.65 mole."},{"Start":"06:41.105 ","End":"06:44.825","Text":"The next step is to calculate the number of moles of hydrochloric acid."},{"Start":"06:44.825 ","End":"06:46.460","Text":"We\u0027re going to do this in the next video."},{"Start":"06:46.460 ","End":"06:48.750","Text":"Thank you very much for watching."}],"ID":23699},{"Watched":false,"Name":"Exercise 1 - Part b","Duration":"3m 48s","ChapterTopicVideoID":22879,"CourseChapterTopicPlaylistID":86818,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22879.jpeg","UploadDate":"2020-12-15T06:11:18.5530000","DurationForVideoObject":"PT3M48S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.355","Text":"Hi. We are going to solve the following exercise."},{"Start":"00:02.355 ","End":"00:04.470","Text":"In the last video, we calculated the number of"},{"Start":"00:04.470 ","End":"00:07.335","Text":"moles of magnesium hydroxide and magnesium carbonate."},{"Start":"00:07.335 ","End":"00:12.165","Text":"The number of moles of magnesium hydroxide equals 4.73 moles,"},{"Start":"00:12.165 ","End":"00:17.490","Text":"and the number of moles of magnesium carbonate equals 1.65 moles."},{"Start":"00:17.490 ","End":"00:21.300","Text":"Now we want to calculate the number of moles of the hydrochloric acid."},{"Start":"00:21.300 ","End":"00:24.360","Text":"Now as you can see, it is present in both equations,"},{"Start":"00:24.360 ","End":"00:26.310","Text":"so we need to use both of them."},{"Start":"00:26.310 ","End":"00:30.225","Text":"Therefore, the number of moles of hydrochloric acid equal."},{"Start":"00:30.225 ","End":"00:32.025","Text":"First of all, look at our first equation,"},{"Start":"00:32.025 ","End":"00:35.415","Text":"the number of moles of magnesium hydroxide times,"},{"Start":"00:35.415 ","End":"00:39.000","Text":"we can see that for every 1 mole of magnesium hydroxide that reacts,"},{"Start":"00:39.000 ","End":"00:43.670","Text":"we have 2 moles of hydrochloric acid which reacts, times 2 moles."},{"Start":"00:43.670 ","End":"00:50.780","Text":"Hydrochloric acid divided by 1 mole of magnesium hydroxide."},{"Start":"00:50.780 ","End":"00:53.420","Text":"That\u0027s our first equation."},{"Start":"00:53.420 ","End":"01:00.720","Text":"Into this, we\u0027re going to add the moles of magnesium carbonate times again,"},{"Start":"01:00.720 ","End":"01:03.560","Text":"for every 1 mole of magnesium carbonate which reacts,"},{"Start":"01:03.560 ","End":"01:06.320","Text":"we have 2 moles of the hydrochloric acid which reacts,"},{"Start":"01:06.320 ","End":"01:10.005","Text":"times 2 moles of hydrochloric acid,"},{"Start":"01:10.005 ","End":"01:15.735","Text":"divided by 1 mole of magnesium carbonate."},{"Start":"01:15.735 ","End":"01:24.620","Text":"This equals 4.73 mole of magnesium hydroxide, which is here,"},{"Start":"01:24.620 ","End":"01:26.900","Text":"which was calculated in a previous video,"},{"Start":"01:26.900 ","End":"01:36.180","Text":"times 2 moles of hydrochloric acid divided by 1 mole of magnesium hydroxide."},{"Start":"01:36.190 ","End":"01:39.950","Text":"The moles of magnesium hydroxide cancel out"},{"Start":"01:39.950 ","End":"01:43.220","Text":"and we have to add to this the number of moles of the magnesium carbonate,"},{"Start":"01:43.220 ","End":"01:45.835","Text":"which is 1.65 mole."},{"Start":"01:45.835 ","End":"01:51.420","Text":"Again, times 2 moles of hydrochloric acid for"},{"Start":"01:51.420 ","End":"01:58.595","Text":"every 1 mole of magnesium carbonate."},{"Start":"01:58.595 ","End":"02:03.125","Text":"The moles of magnesium carbonate cancel out also,"},{"Start":"02:03.125 ","End":"02:07.190","Text":"and this equals 9.46 moles of"},{"Start":"02:07.190 ","End":"02:16.190","Text":"hydrochloric acid plus 3.3 moles hydrochloric acid."},{"Start":"02:16.190 ","End":"02:23.690","Text":"This equals 12.76 moles of hydrochloric acid."},{"Start":"02:23.690 ","End":"02:27.365","Text":"Now you want to calculate the mass of the hydrochloric acid."},{"Start":"02:27.365 ","End":"02:29.630","Text":"In order to calculate the mass of the hydrochloric acid,"},{"Start":"02:29.630 ","End":"02:31.370","Text":"we\u0027ll use the equation n,"},{"Start":"02:31.370 ","End":"02:32.960","Text":"the number of moles equals m,"},{"Start":"02:32.960 ","End":"02:35.880","Text":"which is the mass divided by the molar mass."},{"Start":"02:35.880 ","End":"02:37.760","Text":"Since we\u0027re interested in the mass,"},{"Start":"02:37.760 ","End":"02:41.825","Text":"we\u0027re going to use the equation mass equals the number of moles,"},{"Start":"02:41.825 ","End":"02:43.310","Text":"which we know and we calculated,"},{"Start":"02:43.310 ","End":"02:46.445","Text":"times the molar mass."},{"Start":"02:46.445 ","End":"02:50.285","Text":"So we\u0027re going to calculate the molar mass of hydrochloric acid,"},{"Start":"02:50.285 ","End":"02:52.430","Text":"which is the molar mass of hydrogen,"},{"Start":"02:52.430 ","End":"02:55.010","Text":"plus the molar mass of chlorine,"},{"Start":"02:55.010 ","End":"02:58.790","Text":"which equals 1.01 grams per mole,"},{"Start":"02:58.790 ","End":"03:04.190","Text":"plus 35.45 grams per mole."},{"Start":"03:04.190 ","End":"03:11.410","Text":"This equals 36.46 grams per mole,"},{"Start":"03:11.410 ","End":"03:13.640","Text":"that\u0027s our molar mass."},{"Start":"03:13.640 ","End":"03:16.790","Text":"Now that we have our molar mass and the number of moles,"},{"Start":"03:16.790 ","End":"03:21.320","Text":"you can continue the mass of the hydrochloric acid equals the number of moles,"},{"Start":"03:21.320 ","End":"03:24.190","Text":"which is 12.76 moles,"},{"Start":"03:24.190 ","End":"03:30.595","Text":"times the molar mass, which is 36.46 grams per mole."},{"Start":"03:30.595 ","End":"03:40.045","Text":"Moles cancel out and we get 465.23 grams."},{"Start":"03:40.045 ","End":"03:46.040","Text":"Our final answer is the mass of the hydrochloric acid equals 465.23 grams."},{"Start":"03:46.040 ","End":"03:48.600","Text":"Thank you very much for watching."}],"ID":23700},{"Watched":false,"Name":"Exercise 2","Duration":"8m 20s","ChapterTopicVideoID":22880,"CourseChapterTopicPlaylistID":86818,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22880.jpeg","UploadDate":"2020-12-15T06:11:44.0730000","DurationForVideoObject":"PT8M20S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.700","Text":"We\u0027re going to solve the following exercise."},{"Start":"00:02.700 ","End":"00:06.945","Text":"How many moles of chlorine must be consumed in the first reaction to produce"},{"Start":"00:06.945 ","End":"00:11.730","Text":"3.15 kilograms of dichlorodifluoromethane in the second."},{"Start":"00:11.730 ","End":"00:14.250","Text":"Assume that all the carbontetrachloride"},{"Start":"00:14.250 ","End":"00:17.265","Text":"produced in the first reaction is consumed in the second,"},{"Start":"00:17.265 ","End":"00:22.035","Text":"methane plus chlorine gives us carbontetrachloride plus hydrogenchloride,"},{"Start":"00:22.035 ","End":"00:23.430","Text":"and this is not balanced."},{"Start":"00:23.430 ","End":"00:26.760","Text":"Carbontetrachloride plus hydrogenfluoride gives us"},{"Start":"00:26.760 ","End":"00:30.512","Text":"dichlorodifluoromethane plus hydrogenchloride,"},{"Start":"00:30.512 ","End":"00:31.845","Text":"and this is not balanced."},{"Start":"00:31.845 ","End":"00:35.610","Text":"The first step is to balance the reactions."},{"Start":"00:35.610 ","End":"00:38.760","Text":"Let\u0027s write the first one down methane plus"},{"Start":"00:38.760 ","End":"00:43.560","Text":"chlorine gives us carbontetrachloride plus hydrogenchloride."},{"Start":"00:43.560 ","End":"00:46.790","Text":"Which are going to divide this up into the reactants and the products,"},{"Start":"00:46.790 ","End":"00:50.520","Text":"and then the reactants we can see that we have 1 carbon,"},{"Start":"00:50.660 ","End":"00:57.254","Text":"2 chlorines right here and 4 hydrogens."},{"Start":"00:57.254 ","End":"01:01.305","Text":"In the products we can see that we also have 1 carbon,"},{"Start":"01:01.305 ","End":"01:05.940","Text":"we have 5 chlorines and we have 1 hydrogen."},{"Start":"01:05.940 ","End":"01:07.820","Text":"First of all if we look at our carbons there are"},{"Start":"01:07.820 ","End":"01:10.190","Text":"balanced and the chlorine is and hydrogens are not."},{"Start":"01:10.190 ","End":"01:12.870","Text":"However, we have free chlorine right here,"},{"Start":"01:12.870 ","End":"01:16.590","Text":"so it\u0027s going to be very easy to balance the chlorine is at the end."},{"Start":"01:16.590 ","End":"01:18.030","Text":"Let\u0027s get to the hydrogens."},{"Start":"01:18.030 ","End":"01:19.860","Text":"In our reactants we have 4,"},{"Start":"01:19.860 ","End":"01:21.885","Text":"in our products we have 1."},{"Start":"01:21.885 ","End":"01:25.620","Text":"We\u0027re going to multiply hydrogen chloride by 4 that way you\u0027re"},{"Start":"01:25.620 ","End":"01:28.995","Text":"going to change the hydrogens to 4 hydrogens,"},{"Start":"01:28.995 ","End":"01:33.075","Text":"and now we have 4 chlorines from our hydrogen chloride,"},{"Start":"01:33.075 ","End":"01:37.245","Text":"and 4 more chlorines from our carbontetrachloride which is 8 chlorine,"},{"Start":"01:37.245 ","End":"01:40.030","Text":"so you\u0027re going to change the chlorine to 8."},{"Start":"01:40.120 ","End":"01:43.475","Text":"Carbons are balanced, the hydrogens are balanced,"},{"Start":"01:43.475 ","End":"01:44.510","Text":"and chlorine are not."},{"Start":"01:44.510 ","End":"01:49.350","Text":"We have 8 chlorine in the product and 2 chlorines in the reactants,"},{"Start":"01:49.350 ","End":"01:52.260","Text":"therefore, we\u0027re going to multiply our chlorine by 4,"},{"Start":"01:52.260 ","End":"01:55.365","Text":"and then we\u0027re going to have 8 chlorines because it\u0027s 2 times 4."},{"Start":"01:55.365 ","End":"01:57.765","Text":"We\u0027re going to change the chlorines here from 2."},{"Start":"01:57.765 ","End":"02:01.290","Text":"Now we can see that we have 8 chlorines because it\u0027s 2 times 4."},{"Start":"02:01.290 ","End":"02:04.385","Text":"We\u0027re going to change a chlorine here from 2-8."},{"Start":"02:04.385 ","End":"02:06.020","Text":"Now we have 1 carbon,"},{"Start":"02:06.020 ","End":"02:08.299","Text":"8 chlorines and 4 hydrogens,"},{"Start":"02:08.299 ","End":"02:10.025","Text":"and the reaction is balanced."},{"Start":"02:10.025 ","End":"02:11.750","Text":"We balanced the first equation."},{"Start":"02:11.750 ","End":"02:14.030","Text":"Now we have to balance the second equation."},{"Start":"02:14.030 ","End":"02:19.220","Text":"2 carbontetrachloride plus hydrogenfluoride gives"},{"Start":"02:19.220 ","End":"02:24.905","Text":"us dichlorodifluoromethane plus hydrogenchloride."},{"Start":"02:24.905 ","End":"02:28.925","Text":"We\u0027re going to divide this into the reactants on the product side,"},{"Start":"02:28.925 ","End":"02:31.985","Text":"in the reactants we have 1 carbon,"},{"Start":"02:31.985 ","End":"02:40.570","Text":"4 chlorines, we have 1 fluorine and 1 hydrogen."},{"Start":"02:40.850 ","End":"02:45.915","Text":"Now in the product side we have 1 carbon you can see here we have"},{"Start":"02:45.915 ","End":"02:51.210","Text":"3 chlorines we can see 1, 2, 3."},{"Start":"02:51.210 ","End":"02:55.155","Text":"2 from the dichlorodifluoromethane and 1 from the hydrogenchloride."},{"Start":"02:55.155 ","End":"03:02.050","Text":"We have 2 fluorine and we have 1 hydrogen."},{"Start":"03:04.700 ","End":"03:08.730","Text":"As we can see the carbons are balanced,"},{"Start":"03:08.730 ","End":"03:10.595","Text":"now we go to the chlorine."},{"Start":"03:10.595 ","End":"03:15.020","Text":"We can see that we have 4 on the reactant side and 3 on the product side,"},{"Start":"03:15.020 ","End":"03:18.200","Text":"so we want to change the chlorine to 4."},{"Start":"03:18.200 ","End":"03:20.480","Text":"As we can see we have two chlorines in"},{"Start":"03:20.480 ","End":"03:24.950","Text":"our first molecule and we have 1 chlorine in a second one."},{"Start":"03:24.950 ","End":"03:27.620","Text":"We\u0027re going to multiply or hydrogen chloride by 2"},{"Start":"03:27.620 ","End":"03:30.235","Text":"that way we\u0027ll get 2 chlorines from the hydrogenchloride,"},{"Start":"03:30.235 ","End":"03:34.350","Text":"2 from the dichlorodifluoromethane and we\u0027ll have 4 in all."},{"Start":"03:34.350 ","End":"03:36.320","Text":"We\u0027re going to change the chlorine number from"},{"Start":"03:36.320 ","End":"03:40.235","Text":"3-4 and our hydrogen is also changed from 1-2."},{"Start":"03:40.235 ","End":"03:43.360","Text":"We\u0027re going to change our hydrogens from 1-2."},{"Start":"03:43.360 ","End":"03:46.805","Text":"Now carbons are balanced our chlorine is balanced,"},{"Start":"03:46.805 ","End":"03:49.265","Text":"and the fluorine and the hydrogen is not."},{"Start":"03:49.265 ","End":"03:51.470","Text":"We have 2 fluorines and 2 hydrogens in"},{"Start":"03:51.470 ","End":"03:55.265","Text":"our product side and 1 fluorine and 1 hydrogen in reactant side."},{"Start":"03:55.265 ","End":"03:58.940","Text":"We can fix this very quickly by multiplying our hydrogen fluoride by"},{"Start":"03:58.940 ","End":"04:04.465","Text":"2 that will change the 4 into 2 and our hydrogen and also to 2."},{"Start":"04:04.465 ","End":"04:07.320","Text":"Hydrogen 2 fluoride 2."},{"Start":"04:07.320 ","End":"04:11.175","Text":"Now our carbons are balanced our chlorine is balanced,"},{"Start":"04:11.175 ","End":"04:13.890","Text":"our fluoride is balanced and our hydrogen is balanced,"},{"Start":"04:13.890 ","End":"04:15.990","Text":"so reaction is balanced."},{"Start":"04:15.990 ","End":"04:19.175","Text":"Now we\u0027re going to go on. You want to find the number of moles"},{"Start":"04:19.175 ","End":"04:21.875","Text":"of chlorine reaction and we\u0027re given"},{"Start":"04:21.875 ","End":"04:27.770","Text":"the mass of the dichlorodifluoromethane which is produced in the second reaction."},{"Start":"04:27.770 ","End":"04:29.780","Text":"It\u0027s hard to find the number of moles of the chlorine,"},{"Start":"04:29.780 ","End":"04:31.340","Text":"we\u0027re going to first find the number of moles of"},{"Start":"04:31.340 ","End":"04:35.210","Text":"the dichlorodifluoromethane and then find the number of moles of the chlorine."},{"Start":"04:35.210 ","End":"04:39.890","Text":"The number of moles n equals m which is the mass divided by the molar mass."},{"Start":"04:39.890 ","End":"04:44.810","Text":"The mass of the dichlorodifluoromethane is given and equals 3.15 kilograms the mass"},{"Start":"04:44.810 ","End":"04:50.130","Text":"of dichlorodifluoromethane equals 3.15 kilograms."},{"Start":"04:50.130 ","End":"04:54.630","Text":"We\u0027re going to calculate the molar mass of dichlorodifluoromethane."},{"Start":"04:54.970 ","End":"05:01.970","Text":"This equals the molar mass of carbon plus 2 times the molar mass of chlorine since we"},{"Start":"05:01.970 ","End":"05:03.770","Text":"have two chlorine atoms plus"},{"Start":"05:03.770 ","End":"05:08.300","Text":"2 times the molar mass of fluorine since we have two fluorine atoms in the molecule."},{"Start":"05:08.300 ","End":"05:15.215","Text":"This equals 12.01 grams per mole plus 2 times 35.45 grams per mole,"},{"Start":"05:15.215 ","End":"05:17.555","Text":"plus 2 times 19 grams per mole,"},{"Start":"05:17.555 ","End":"05:20.855","Text":"which equals 120.91 grams per mole."},{"Start":"05:20.855 ","End":"05:24.650","Text":"Now the molar masses were taken from the periodic table of elements."},{"Start":"05:24.650 ","End":"05:28.100","Text":"Now we\u0027re going to calculate the number of moles of dichlorodifluoromethane."},{"Start":"05:28.100 ","End":"05:32.850","Text":"The number of moles of dichlorodifluoromethane equals the mass of"},{"Start":"05:32.850 ","End":"05:39.305","Text":"dichlorodifluoromethane divided by the molar mass of dichlorodifluoromethane."},{"Start":"05:39.305 ","End":"05:43.140","Text":"This equals 3.15 kilograms"},{"Start":"05:44.380 ","End":"05:50.765","Text":"divided by 120.91 grams per mole,"},{"Start":"05:50.765 ","End":"05:52.910","Text":"which we just calculated."},{"Start":"05:52.910 ","End":"05:58.080","Text":"Now here we have kilograms at our numerator and then denominator we have grams per mole,"},{"Start":"05:58.080 ","End":"06:00.050","Text":"so we\u0027re going to multiply the numerator by"},{"Start":"06:00.050 ","End":"06:03.980","Text":"a conversion factor in every one kilogram we have 1,000 grams,"},{"Start":"06:03.980 ","End":"06:06.500","Text":"so it\u0027s 1000 grams per one kilogram."},{"Start":"06:06.500 ","End":"06:09.290","Text":"Kilograms cancel out and we\u0027re left with grams,"},{"Start":"06:09.290 ","End":"06:13.675","Text":"so this equals 26.05 mole."},{"Start":"06:13.675 ","End":"06:16.520","Text":"Remember that when you divide by a fraction is the same"},{"Start":"06:16.520 ","End":"06:19.490","Text":"as multiplying by the reciprocal of the fraction,"},{"Start":"06:19.490 ","End":"06:23.105","Text":"in the units we have grams divided by grams per mole,"},{"Start":"06:23.105 ","End":"06:27.105","Text":"so this equals grams times mole."},{"Start":"06:27.105 ","End":"06:30.550","Text":"The grams cancel out and we\u0027re left with moles."},{"Start":"06:30.550 ","End":"06:36.210","Text":"We know that the moles of the dichlorodifluoromethane equal 26.05 mole,"},{"Start":"06:36.210 ","End":"06:40.325","Text":"now we want to find and calculate the moles of chlorine in the first reaction"},{"Start":"06:40.325 ","End":"06:42.980","Text":"and we also know that all the carbontetrachloride"},{"Start":"06:42.980 ","End":"06:44.900","Text":"which is produced in the first reaction,"},{"Start":"06:44.900 ","End":"06:46.480","Text":"is consumed in a second."},{"Start":"06:46.480 ","End":"06:48.365","Text":"If we look at the second reaction,"},{"Start":"06:48.365 ","End":"06:54.154","Text":"we can see that the number of moles of dichlorodifluoromethane which are produced,"},{"Start":"06:54.154 ","End":"06:56.030","Text":"equals the number of moles of"},{"Start":"06:56.030 ","End":"07:00.575","Text":"carbontetrachloride which are consumed and if we look at our first reaction,"},{"Start":"07:00.575 ","End":"07:03.515","Text":"you can see the most of the carbontetrachloride which are"},{"Start":"07:03.515 ","End":"07:08.115","Text":"produced is less than 4 times the number of chlorine moles."},{"Start":"07:08.115 ","End":"07:09.620","Text":"If we look at our first reaction,"},{"Start":"07:09.620 ","End":"07:13.260","Text":"we can see that for every 4 moles of chlorine which react,"},{"Start":"07:13.280 ","End":"07:17.650","Text":"we get 1 mole of carbontetrachloride."},{"Start":"07:17.650 ","End":"07:19.250","Text":"Now if we look at our reactions,"},{"Start":"07:19.250 ","End":"07:23.065","Text":"we can see that in the first reaction for every 4 moles of chlorine which react,"},{"Start":"07:23.065 ","End":"07:26.790","Text":"we get 1 carbontetrachloride mole and if we look at"},{"Start":"07:26.790 ","End":"07:31.175","Text":"our second reaction we can see that for every 1 mole of carbontetrachloride which react,"},{"Start":"07:31.175 ","End":"07:34.700","Text":"we get 1 mole of dichlorodifluoromethane meaning that for"},{"Start":"07:34.700 ","End":"07:39.510","Text":"every 1 mole of chlorine which reacts with 1 mole of dichlorodifluoromethane."},{"Start":"07:39.510 ","End":"07:43.440","Text":"The number of moles of chlorine equal the moles of"},{"Start":"07:43.440 ","End":"07:46.830","Text":"the dichlorodifluoromethane times 4 moles of"},{"Start":"07:46.830 ","End":"07:52.395","Text":"chlorine divided by 1 mole of dichlorodifluoromethane."},{"Start":"07:52.395 ","End":"07:55.665","Text":"This equals 26.05 mole of"},{"Start":"07:55.665 ","End":"07:58.590","Text":"dichlorodifluoromethane times 4 moles of"},{"Start":"07:58.590 ","End":"08:03.285","Text":"chlorine divided by 1 mole of dichlorodifluoromethane."},{"Start":"08:03.285 ","End":"08:12.705","Text":"The moles of dichlorodifluoromethane cancel out and we get 104.2 moles of chlorine."},{"Start":"08:12.705 ","End":"08:16.610","Text":"The number of moles of chlorine equal 104.2 mole."},{"Start":"08:16.610 ","End":"08:18.185","Text":"That is our final answer."},{"Start":"08:18.185 ","End":"08:20.580","Text":"Thank you very much for watching."}],"ID":23701},{"Watched":false,"Name":"Simultaneous Reactions","Duration":"5m 31s","ChapterTopicVideoID":16919,"CourseChapterTopicPlaylistID":86818,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/16919.jpeg","UploadDate":"2019-02-20T23:49:40.8700000","DurationForVideoObject":"PT5M31S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.620","Text":"In the previous video,"},{"Start":"00:01.620 ","End":"00:06.855","Text":"we discussed consecutive reactions where 1 reaction occurs after another."},{"Start":"00:06.855 ","End":"00:09.315","Text":"That is of course consecutively."},{"Start":"00:09.315 ","End":"00:12.945","Text":"In this video, we will discuss simultaneous reactions,"},{"Start":"00:12.945 ","End":"00:17.040","Text":"which occur, not surprisingly, simultaneously."},{"Start":"00:17.040 ","End":"00:20.820","Text":"Simultaneous reactions are 2 or more reactions"},{"Start":"00:20.820 ","End":"00:24.210","Text":"that occur at the same time in the same system."},{"Start":"00:24.210 ","End":"00:25.919","Text":"Let\u0027s take an example."},{"Start":"00:25.919 ","End":"00:34.050","Text":"5 grams of MnO_2 and 5 grams of Mn_3O_4 react with excess CO,"},{"Start":"00:34.050 ","End":"00:35.745","Text":"that\u0027s Carbon monoxide,"},{"Start":"00:35.745 ","End":"00:39.915","Text":"to form MnO and CO_2 carbon dioxide."},{"Start":"00:39.915 ","End":"00:43.520","Text":"Calculate the mass of CO_2 formed."},{"Start":"00:43.520 ","End":"00:48.155","Text":"We\u0027ll begin by writing balanced equations for each reaction."},{"Start":"00:48.155 ","End":"00:50.180","Text":"Here\u0027s the first reaction,"},{"Start":"00:50.180 ","End":"00:52.115","Text":"MnO_2 plus CO,"},{"Start":"00:52.115 ","End":"00:54.695","Text":"giving MnO plus CO_2."},{"Start":"00:54.695 ","End":"00:58.120","Text":"Let\u0027s check whether this equation is balanced."},{"Start":"00:58.120 ","End":"01:00.330","Text":"We\u0027ve 1 Mn on the left-hand side,"},{"Start":"01:00.330 ","End":"01:01.920","Text":"and 1 on the right-hand side."},{"Start":"01:01.920 ","End":"01:05.890","Text":"1 Carbon on the left and 1 Carbon on the right."},{"Start":"01:05.890 ","End":"01:09.560","Text":"2 plus 1 Oxygens on the left,"},{"Start":"01:09.560 ","End":"01:12.365","Text":"and 1 plus 2 Oxygens on the right."},{"Start":"01:12.365 ","End":"01:15.500","Text":"We have 3 on each side, 3 to 3."},{"Start":"01:15.500 ","End":"01:18.305","Text":"This equation is balanced."},{"Start":"01:18.305 ","End":"01:20.405","Text":"Let\u0027s take the next equation."},{"Start":"01:20.405 ","End":"01:24.125","Text":"We have Mn_3O_4 plus CO,"},{"Start":"01:24.125 ","End":"01:26.870","Text":"giving MnO plus CO_2."},{"Start":"01:26.870 ","End":"01:30.355","Text":"Let\u0027s look at the Manganese Mn_3."},{"Start":"01:30.355 ","End":"01:32.145","Text":"We have 3 on the left,"},{"Start":"01:32.145 ","End":"01:33.825","Text":"but only 1 on the right,"},{"Start":"01:33.825 ","End":"01:35.820","Text":"so we have to multiply by 3."},{"Start":"01:35.820 ","End":"01:37.290","Text":"Let\u0027s look at the Carbons,"},{"Start":"01:37.290 ","End":"01:40.555","Text":"we\u0027ve 1 on the left and 1 on the right. That\u0027s okay."},{"Start":"01:40.555 ","End":"01:42.700","Text":"Now look at the Oxygens,"},{"Start":"01:42.700 ","End":"01:45.750","Text":"4 plus 1, that\u0027s 5,"},{"Start":"01:45.750 ","End":"01:47.385","Text":"and the right-hand side,"},{"Start":"01:47.385 ","End":"01:49.455","Text":"3 times 1, is 3,"},{"Start":"01:49.455 ","End":"01:53.540","Text":"plus 2 in carbon dioxide giving us a total of 5."},{"Start":"01:53.540 ","End":"01:56.215","Text":"This equation is balanced."},{"Start":"01:56.215 ","End":"01:59.695","Text":"Now we can write the scheme for solving the problem."},{"Start":"01:59.695 ","End":"02:04.885","Text":"We\u0027re given the mass of the reactants of MnO_2 and Mn_3O_4."},{"Start":"02:04.885 ","End":"02:08.260","Text":"We have to calculate the moles of reactants."},{"Start":"02:08.260 ","End":"02:12.075","Text":"Then from that to the moles of Carbon dioxide,"},{"Start":"02:12.075 ","End":"02:14.850","Text":"and from that to the mass of Carbon dioxide."},{"Start":"02:14.850 ","End":"02:19.130","Text":"We\u0027ll call these steps 2, 3 and 4."},{"Start":"02:19.130 ","End":"02:20.750","Text":"Here\u0027s step 2,"},{"Start":"02:20.750 ","End":"02:23.930","Text":"the masses of reactants to most of reactors."},{"Start":"02:23.930 ","End":"02:28.970","Text":"The molar mass of MnO_2 is 86.94 grams per mole."},{"Start":"02:28.970 ","End":"02:32.105","Text":"From that, we can calculate the number of moles."},{"Start":"02:32.105 ","End":"02:37.835","Text":"Now we know that the number of moles is equal to mass divided by the molar mass."},{"Start":"02:37.835 ","End":"02:41.480","Text":"We have the number of moles of MnO_2, is the mass,"},{"Start":"02:41.480 ","End":"02:45.335","Text":"5 grams divide by 86.94 grams per mole,"},{"Start":"02:45.335 ","End":"02:46.960","Text":"which is the molar mass."},{"Start":"02:46.960 ","End":"02:48.240","Text":"Divide the 2 numbers,"},{"Start":"02:48.240 ","End":"02:51.405","Text":"we get 0.05751,"},{"Start":"02:51.405 ","End":"02:56.555","Text":"grams cancels with grams and 1 over molar minus 1 is mole."},{"Start":"02:56.555 ","End":"03:00.455","Text":"Now we have the number of moles of MnO_2."},{"Start":"03:00.455 ","End":"03:04.715","Text":"Let\u0027s calculate the number of moles of Mn_3O_4."},{"Start":"03:04.715 ","End":"03:07.010","Text":"First of all, we need the molar mass."},{"Start":"03:07.010 ","End":"03:12.060","Text":"That\u0027s 228.8 grams per mole."},{"Start":"03:12.060 ","End":"03:15.430","Text":"I forgot to put grams per mole in."},{"Start":"03:16.340 ","End":"03:19.350","Text":"The number of moles is its mass,"},{"Start":"03:19.350 ","End":"03:27.930","Text":"5.00 divided 228.8 grams per mole."},{"Start":"03:27.930 ","End":"03:30.060","Text":"We divide the 2 numbers,"},{"Start":"03:30.060 ","End":"03:33.270","Text":"we get 0.02185,"},{"Start":"03:33.270 ","End":"03:39.835","Text":"and the grams cancels with grams and 1 over molar minus 1 gives us mole."},{"Start":"03:39.835 ","End":"03:46.295","Text":"Now we know the number of moles of Mn_3O_4."},{"Start":"03:46.295 ","End":"03:47.780","Text":"Now step 3,"},{"Start":"03:47.780 ","End":"03:52.010","Text":"we want to go from the moles of reactants to the moles of CO_2."},{"Start":"03:52.010 ","End":"03:53.730","Text":"If we look at the 2 equations,"},{"Start":"03:53.730 ","End":"03:59.920","Text":"we see that 1 mole of CO2 is produced for 1 mole of Manganese Oxide."},{"Start":"03:59.920 ","End":"04:02.580","Text":"Happens for both equations,"},{"Start":"04:02.580 ","End":"04:08.265","Text":"for MnO_2 and for the 1 for Mn_3O_4."},{"Start":"04:08.265 ","End":"04:11.150","Text":"For that, we can write that the number of moles of"},{"Start":"04:11.150 ","End":"04:14.330","Text":"Carbon dioxide is exactly equal to the number of"},{"Start":"04:14.330 ","End":"04:21.580","Text":"moles of MnO_2 plus the number of moles of Mn_3O_4,"},{"Start":"04:21.580 ","End":"04:28.295","Text":"and we can add these 2 numbers and get 0.07936 moles."},{"Start":"04:28.295 ","End":"04:31.595","Text":"That\u0027s the number of moles of Carbon dioxide"},{"Start":"04:31.595 ","End":"04:37.065","Text":"produced from the 1st and from the 2nd reaction."},{"Start":"04:37.065 ","End":"04:40.285","Text":"The final step, step 4,"},{"Start":"04:40.285 ","End":"04:45.455","Text":"is to calculate the mass of Carbon dioxide from the number of moles."},{"Start":"04:45.455 ","End":"04:50.565","Text":"The molar mass of Carbon dioxide is 44.01 grams per mole."},{"Start":"04:50.565 ","End":"04:54.530","Text":"The mass of Carbon dioxide is a number of moles,"},{"Start":"04:54.530 ","End":"04:59.715","Text":"0.07936 times the molar mass,"},{"Start":"04:59.715 ","End":"05:03.155","Text":"44.01 grams per mole."},{"Start":"05:03.155 ","End":"05:05.060","Text":"We multiply these 2 numbers,"},{"Start":"05:05.060 ","End":"05:07.655","Text":"we get 3.493,"},{"Start":"05:07.655 ","End":"05:13.730","Text":"and the units are just grams because moles times moles to minus 1 is 1."},{"Start":"05:13.730 ","End":"05:16.254","Text":"So we\u0027re left just with grams."},{"Start":"05:16.254 ","End":"05:21.285","Text":"Here\u0027s our final answer, 3.493 grams."},{"Start":"05:21.285 ","End":"05:22.520","Text":"So in this video,"},{"Start":"05:22.520 ","End":"05:26.465","Text":"we discussed simultaneous reactions."},{"Start":"05:26.465 ","End":"05:29.615","Text":"We\u0027ve got CO2 from each reaction,"},{"Start":"05:29.615 ","End":"05:31.830","Text":"and we added the 2."}],"ID":17666}],"Thumbnail":null,"ID":86818}]

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