Lecture 14. Limiting Reagents
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Lecture/Conference
Transcript: English(auto-generated)
00:13
It is so nice to see you. And I prepared for you, again, a wonderful menu of chemistry items.
00:29
The title is nothing less than limited reagents. So before I do that, I'm going to dive into some of what we talked about last time.
00:44
The last item is basically if you have a chemical reaction, and you're given a certain amount of the starting material, you can calculate how many grams of the product you generate. There's a couple of conversions there, and we're going to come back to that right now.
01:02
The basics of this is all, you know, I have to put this picture up, it's all counting. This is all about counting numbers of molecules and compounds. How many do you have of something? And you may convert the grams, the mass, to countable objects, which
01:21
we happen to express in moles. Moles are just the mean quantity for expressing the number of molecules in it. So that's how we use the mole. Now, this is a question that we had last time, we solved it, okay, and the way we did this, the very last slide,
01:43
just to remind you, we started with the starting material, we had one gram, we made a conversion from grams to moles. Because moles is a quantity, a countable object. Now we can actually compare two quantities from another.
02:01
I cannot compare two masses from another because every molecule has its own mass. That's the same, as I say, that if I deal with you guys, all right, you guys have all different kinds of masses. So a mass of your neighbor may not be the same as your own mass, so dealing with comparing masses doesn't necessarily tell me anything about
02:24
how many people are in this audience. Right? If I want to do that, I have to go away from mass to countable objects. The mass of a human being is not necessarily important in that regard. So I'm going from mass to a
02:44
countable object, which is moles, and then I can relate one quantity to another quantity. The quantity of starting material is relative to the amount of the product. The ratio is one to one, that's what the chemical equation says. So each one mole of this, you generate one mole of the other, and then you go from moles, which
03:07
you now have, the product is now expressed in moles, I want to express the product. In grams, again, I make this conversion. From grams to moles, mole ratio, and then from moles to grams. And we found the answer. The same answer last time, of course, 72.3 grams.
03:23
And this falls for, you know, any kind of chemical equation. So that means it also falls for this one. Yes, first one. Say again. This is the
03:42
molar mass of this compound. How do you get that? You go to your table, two times chlorine, and you add them all up. Okay. So let's quickly do this for another question and see if we get this. So this has solid zinc reacts with hydrochloric acid, that's here, those are the reactions,
04:01
and it produces two products, namely zinc chloride and hydrogen gas. Right there. The question is, what mass of zinc chloride, this product, do you form if you have 1.528 grams of solid zinc? And we assume then that this is an excess. So this is not
04:20
a limiting factor in the calculation. There is enough HCl such that all of this will be converted into that. So how do you do this? Well, you start with the amount of starting material, again, 1.5 grams of zinc quickly convert that to the number of moles. Always going to move away from moles, go through moles, that's where everything happens.
04:44
So that's what we do, how do we do that? Well, we divide by the molar mass. The molar mass of zinc is what you get from the periodic level. Then, I have my quantity in the number of zinc atoms, expressed in moles, and then I can see, okay, for each one zinc atom, I generate one zinc chloride compound. So it's one-to-one. So
05:08
that means I can convert from zinc to zinc chloride. One-to-one. But this doesn't have to be moles to one. I mean, you can have a chemical equation where this is four and this is five, and then actually this is going to be five and this is four. Right? And we'll see an example of that. So, now I have converted this quantity into the quantity, the number
05:29
of zinc chloride units, and now I'm going to convert that back to the number of grams. So what I'm doing then, I'm going to multiply the number of zinc chloride units, expressed in moles, with the molar mass to find out how many grams of that stuff I have.
05:47
Okay? In this case you find 3.185 grams of zinc chloride. This is exactly the same calculation, different days, but this is a general calculation. There's two things you can do. You can try to put this in your head, and memorize the method, and then try to recognize where
06:06
you want to apply this, and then hopefully you're right when you do that. I'd rather have you not do that. Because that's the kind of rule that dissipates. You can do your exam and it's completely gone. What I'd rather have is that you try to get a feel for this.
06:27
Why do you do this calculation of radiation when you do it? If you understand, you can reproduce that, you can explain to yourself why you do it this way, that way. Even better, if you can explain it to your four-year-old sister, that's even better. Okay. So let's move on to the following situation.
06:47
You look at chemical equations, you cannot calculate, given the amount of starting material, how much product will it carry away. But if you don't look at it, it's a very important and very real situation. Namely, when does the
07:01
reaction just stop? When does it really stop? And it does stop intuitively when both of the reactants run out. So let's say if I have this chemical here, which is a chemical reaction, this wax. The wax is being combusted, which means it reacts with oxygen. There it is. This is wax.
07:25
Nice hydropower. And it reacts with oxygen to produce, again, H2O in the property of the oxide. That's the combustion reaction. You've seen many of these now. When does this reaction stop? Well, it stops when one of the reactants is running out.
07:47
So if I have no more wax, of no more oxygen, the reaction can't go on. That's when it stops. However, the question is, which of these reactants is running out first?
08:03
Well, in the case of this particular candle, or in general, if you like a candle, when does the candle stop? When the oxygen is run out, or when the wax is gone? Typically, when the wax is gone. This is the sum of oxygen. Only if you put a glass table around it, that's when you lose the oxygen. But if you just have a glass table around it, that's when you lose the oxygen.
08:24
There's so much oxygen around, this oxygen here is not going to be effective in this combustion reaction. There's enough of it. When it comes to oxygen, this thing will go on forever. That's why if you have a forest fire, the forest fire will just go on. So it's the other part, and there's the wax. So the wax in this
08:47
thing is called the limiting reagent, and that is something we've talked about right now. The limiting reagent is that reagent that's running out first, and hence will stop the reaction.
09:01
And because it's running out first, it will also determine how much product you will form. So if you have to calculate how much CO2 you will form, it doesn't make a lot of sense to look at how much oxygen you have in the beginning, because there's enough of it. It is determined by this guy. This guy is defining how much of this CO2 you will form. The limiting reagent determines, basically, what is the
09:24
the number of molecules you produce. All right. So just to visualize that, okay, you have to make it simple for yourself, here is a very simple chemical reaction. Mainly, the blue guy interacts with the red girl, and they form a color.
09:46
Now, let's say we have three blue boys, and four red girls. How many purple can you make? That's right. Two colors. One, two, three.
10:03
Well, I have also one red girl left. She can't find a partner. Very, very sad. So what is the limiting reagent? The blue guy is the limiting reagent, and the girls are excess. They are excess.
10:24
There's too many girls. So we call the red girls here, in this case, the excess reagent. And the same rules for molecules, of course, right? Exactly the same, there's absolutely no difference. And even these numbers, you think, oh, it's just like three or four, that's easy, hey, three or four moles, okay, things like that are also easy. It's the same. There's no difference.
10:48
Okay. So the limiting reagent runs at first, limits the amount of product you can form. Okay? This guy here, one, two, three, determines how many couples you can make. The girls don't determine that because there's enough. But this guy determines how many couples you can make. It defines how many molecules you
11:05
can form. The rest is in excess and is left over after the reaction has come from the soft. the excess radiation, and we are able to interpret how much we have left as well. So we'll do that in the next couple of slides.
11:28
Okay. Now let's do a quick exercise here. Here's a chemical reaction, nitrogen interacts with hydrogen gas to form ammonia molecules.
11:43
So let's again do this in a visual way. Instead of having the girls and boys, we have molecules now. For some people, this is too much of a rule to take. You place the girls and the girls with the word molecule and suddenly they don't use that name anymore. So don't let it be you, okay? What I'm trying to do is show exactly the same thing. There's no difference.
12:06
Molecules are the same as all of us. Here it is. Two nitrogen molecules, three hydrogen molecules, I have two hydrogen molecules. This is my starting situation. How many of these guys can I form? Now these guys are now composed of one nitrogen, so each of these molecules needs one nitrogen, it needs three hydrogens.
12:29
I have four nitrogen atoms, I have one, two, three, four, I have six hydrogens, six hydrogens means I can form two of these molecules, and then I'm running out of hydrogens.
12:41
So what I can do is I can take one, two, six, all six, combining the two nitrogen atoms, then I can make these two guys. One, two, three, one, two, three, that's six, six here, one, two blue guys, blue guys here. So, in other words, I have completed my reaction
13:03
I have one left, that is apparently the one that's in excess. So in the same kind of terminology as the previous slide, you can call this the limiting reagent. The hydrogen molecules are running out first, they determine how much ammonia can form. And the other guy is the excess reagent. In this case, the nitrogen molecule.
13:29
Okay. Now, let's not look at a question like this. Let's determine what is the limiting reagent in this particular question. All right. So I have a mixture of magnesium and iodine, iodine gas, I have 5
13:46
grams of magnesium and I have 75 grams of iodine gas. And my product is called magnesium iodide. Identify the limiting reagent. Okay? So which one is running out first in this particular case?
14:01
Now if you look at the masses, you may argue, oh, look, I have 5 grams of magnesium and 75 of iodine. So this one must run out first because I have less of it. That is not how it works. Why not? You don't know how much this guy weighs. I mean, what if this guy, his mass is 5 grams? And he only won.
14:23
And what if this guy weighs like a fraction of that? Or vice versa. You don't really know how many grams of magnesium he weighs. How many of these magnesium atoms do you have? How many iodine molecules do you have? So you really have to convert these numbers here to countable objects. Numbers. Okay? So that's what we're going to do.
14:46
Before I do that, I want to check very quickly if I have one mole of magnesium, how many moles of this do I need to make a product? That's called the mole ratio. Okay? These guys like to react one-to-one.
15:02
One mole of magnesium and one mole of iodine gas will react one-to-one, and they'll exactly form one mole of magnesium iodine. So if I have the exact same quantity, the exact one mole each, they run out at the same time. And there's no limitation. They both have no limitation. But if one of them is more or less, then you have a little variation.
15:22
Okay? So this is the mole ratio of the balanced equation. The balanced equation says these guys react one-to-one. Now let's see what the actual mole ratio is. What is the actual mole ratio? It is how much you actually have.
15:41
What I actually have is this and that, so I'm going to convert these numbers to number of moles. So, 5 grams of magnesium I'm going to convert it to number of moles of magnesium. I do that by dividing by the molar mass. 24.3 is the molar mass of magnesium
16:02
and I find a total of .2 moles of magnesium. And I do the same for iodine. There it is, 75 grams of iodine. Now, note, this is iodine 2. So that means you multiply the molar mass of the iodine atom by 2, right? So 1 mole of iodine 2 is this much. Twice the molar mass of the iodine atom is the molar mass of iodine 2.
16:30
And you find this much. So .269 moles of iodine 2. 96. Thank you. And, if you take the ratio of that, so I do exactly the same as here.
16:43
1 mole of magnesium, 1 mole of iodine, I have this many moles of magnesium and this many moles of iodine, that ratio is .697. So why do I do this? Well, this is more than that. Magnesium's on top, iodine's on the bottom, magnesium's on top, iodine's on the bottom.
17:02
It's less than this number. What does that mean? That means apparently, I don't have enough magnesium to make this one. Okay? If this number would be larger, I get closer to 1, it would be really large, it would be more than 1. So basically, if this is less than 1, that means that the numerator is a little bit big. I'm running out of magnesium first, because there's less of
17:27
it. Yes. You can, okay, this is a good question, but the question is can I just look at the number of moles here and decide which one
17:41
is more? The answer is yes, you can do that if you have a reaction where it reacts one to one. What are the reactions that you get? It's not like that. It's like 9 to 14. Then you have a harder time to figure it out. Yes? Okay. So let me continue the next slide and I'll tell you exactly this particular method. This is kind of a method, and this is the
18:08
method that always works, and you would be wise to apply it. There's more ways in which you can get to this answer. Okay? But let's see. What you are going to do is this, identify the mole ratio of the balance equation. That was the 1 to 1 we just identified.
18:27
The mole ratio between the reagents. Then we are going to calculate the mole ratio of the reagents and the actual number of moles is half. We use quantities that are given in the question notes.
18:43
You call that the actual mole ratio. and then you're going to compare the mole ratio as stipulated by the equation with the mole ratio that you get from the actual numbers of molecules that you have. And then you have two possibilities. Namely, if the actual mole ratio is smaller
19:05
than a balanced mole ratio, then the numerator is the limiting reaction. So in the previous case, we found the balance was 1, this actual volume is less than 1, which means the numerator is the limiting reaction. And you can also have a reverse situation, if the actual mole ratio
19:23
is larger than the balanced mole ratio, then the denominator is the limiting reaction. So this is really a trick. I don't do light tricks, and in this particular case, it is a trick that will always get you to the right answer.
19:41
So this is just a quick way to determine what you have more of. That's very interesting. Okay? So what we have to do then in all these kind of exercises is to identify the mole ratio from the equation. The equation tells you what the proportion should be, and then react completely, and let's say you both
20:10
quantities of the reagent are perfectly matched. In reality, this is typically not the case. We call that the actual mole ratio. So I prepared the balanced mole ratio for the equation, the actual mole ratio to calculate or get you to
20:25
determine which one of the reagents is the limiting reaction. Okay. Now let's do a quick reaction because it's a lot of power. Okay. I have a reaction of ammonia with oxygen. It forms nitrogen monoxide and water. So this is a reaction, and the
20:43
question is, what mass of nitrogen monoxide forms and how much of a product does that form? And, very important, how many grams of the excess reagent do we have left? Okay? So this is a question, a very difficult question, I have a reaction
21:01
there's a limiting reagent which defines how much product I form, and that also tells you how much of the excess reagent do I have left after the reaction stops. There are two questions there. Two questions in one. Okay. So the first thing that you would like to do is to figure out what kind of chemical reaction am I talking about.
21:25
So I have ammonia, I have oxygen, those are the reagents, nitrogen monoxide, and water. Those are the products. So here is the balanced equation for you. Ammonia, oxygen, nitrogen monoxide, and
21:42
water right there, it has the stoichiometric coefficients. Look. These are not one. Okay? Four, five, four, six. So here we really have to take care of the ratios properly. So now we know the equation, next, I would like to know
22:01
which one of these two reagents, there's two, one of them is the limiting reagent. Okay. So let's do then what we just talked about. I'm going to calculate first the mole ratio between these two guys based on the equation. The equation says you need four moles of ammonia for each five moles, and that's 0.8. Now, why would
22:27
I put nitrogen on top and oxygen on the bottom? It doesn't matter. I can flip this over as well. If I put nitrogen on top and ammonia on the bottom, I would get a different number.
22:40
That number has no particular meaning, it only has meaning relative to what I'm going to put in the parent grid, which is the actual mole ratio. As long as you calculate the mole ratio for the balanced equation, in the same way as the actual mole ratio, which means the same guy on the top, the same guy at the bottom, you'll be fine.
23:01
So it doesn't matter which one you put on top or the bottom, as long as you be consistent in your calculation. Okay. Point A is the one we determined here now we're going to determine what is the actual mole ratio. For that, I need these actual values. In grams, I see, but I don't need grams, I need moles. So I'm going to convert these three grams and five grams
23:21
into moles. By dividing these numbers by their respective molar masses. So there we go. I have three grams of ammonia, I like to convert it into number of moles, so I'm going to divide that by the molar mass of ammonia, to 17 grams per mole.
23:41
That calculation tells me 0.17 moles of ammonia. I have another reagent, that is oxygen. I have five grams of oxygen. Now, again, be careful here, this is O2. A lot of folks tend to forget that these are two oxygens.
24:02
So you have to multiply the molar mass of oxygen atoms with two to get to this molar mass. I divide this by the molar mass of oxygen two, which is 32 grams, so I find 0.156 moles of oxygen moles. Okay. So let's calculate
24:21
the mole ratio, and I do something consistent relative to the previous mole ratio. I put NH3 on top, and oxygen on the bottom. Same as what I did in this slide. And then I'm going to compare these two guys. This is 1.13. 1.13 is more than 0.8. If it's more,
24:44
that means that the denominator is the limiting reagent. Okay? The limiting reagent is O2. That means that the other guy, ammonia, is the excess reagent. Okay? So you will have some ammonia left after the reaction stops.
25:02
Oxygen is completely gone, O reacted, NH3, something's left. And I need to determine both. I need to determine how much product we afford, and how much of this stuff we have left after the reaction. Okay. So let's move on and calculate how much product we afford.
25:23
I need the amount of limiting reagent, because the limiting reagent determines how much product I afford. So what I'm going to do is, I'm going to say, look, I have 0.50 moles of oxygen. Oxygen is the limiting reagent. And that's the number I'm just calculating. Then, I would like to know from each mole of
25:45
oxygen, how many moles of nitrogen and monoxide do I generate? In other words, what is the mole ratio between the two? So I'm going to convert the number of moles of oxygen into the number of moles of product. How? I'm making the mole ratio between the two. 4 moles of the product for each 5 moles of oxygen.
26:03
I make a ratio of oxygen at the bottom, and nitrogen monoxide at the top, because I want to go to nitrogen monoxide. Just like that. Moles of oxygen will strike out, and I'm converting this whole thing into moles of nitrogen monoxide, which is the number of moles of the product.
26:24
I can immediately multiply that by the molar mass of nitrogen monoxide. So, this first part here is the number of moles of nitrogen monoxide, the last conversion is going from moles to grams. I'm multiplying by the molar mass of nitrogen
26:42
monoxide. This calculation tells me I have 3.75 grams of product. You see, these calculations can be fairly quick. They're fairly quick. Okay. So that's the first part. The second part is, how much of the excess of reagent do I have left? How do you do that?
27:07
Well, what I want to know is how much has reacted. And how much did I have left? So I had a whole bunch of ammonium molecules in the beginning, some of them had reacted, and the rest is what
27:24
the reaction was. So how do I determine how much of the ammonium molecules reacted? Well, that's determined, again, by the reaction. By how much limiting reagent you have. The limiting reagent determines how much salt will react. So what I have to do then, I again start the limiting reagent, the amount of it,
27:42
0.50 moles, and I have to convert this into the number of moles of the other reagent. How? The mole ratio. Okay? From each 5 moles of oxygen, I have 4 moles of, or I need, 4 moles of ammonium to react.
28:00
into 4 moles of ammonium oxide. So converting oxygen into ammonia involves this. 4 moles of ammonia feeds 5 moles of oxygen. Moles of oxygen will cross out, and I convert this whole quantity here into moles of an ammonia reactor.
28:22
This is the amount of ammonium that has to react in the reaction. Okay? That's a total of 0.125. So 0.125 moles of ammonia has to react to this. After all the oxygen runs out, this is the amount of ammonia that came with it in the reactor.
28:46
The rest of the ammonia is still in the pot. So what do I do? Well, I take the amount that I had in the beginning, I calculated that from two slides back, I calculated that I had in the beginning, basically, I have 3 grams of ammonia divided by the mole mass was 0.17 moles, remember, but this is what I had at the beginning,
29:07
this is the amount that has reacted, it is gone, so we subtract the amount that has gone from the amount that we have at the beginning, and we get a left. And that is this much. 0.051 moles of ammonia. That is the amount in excess.
29:26
All right. I'm almost there, because they say, how many grams? I have the amount of moles, I multiply it with the mole mass of ammonia, which is 70 grams, or mole, and I find the answer. 0.869 grams.
29:48
So this is a little bit more involved, because you have to realize that the amount of excess reagent to determine that, you have to first calculate how much has reacted, and subtract that from the amount that you had at the beginning. This may seem logical, but I can tell you, honestly, about half of the people on the exam will do this
30:09
wrong. About half of the people will do this wrong. They think that the amount of excess reagent is simply either this or that. It's not.
30:24
Think of it as the boys and girls that we had at the beginning. Everybody saw it directly at this one girl left. Because you subtract 3, which is reactive, from 4. You have 4 in the beginning, 3 as the reactor, and you have 1 left. It's the same thing. Except for the same thing. Don't be mistaken.
30:45
Every year, 50% of you guys will not do this. Just forget the last step. Yes?
31:02
Well, I mean, it depends on the kind of equation. If the equation is simple enough, if you can balance it yourself, that could be the first part of the equation could be balance this equation. And then you continue the calculation with the mole ratio in the cell. So yes, you should be able to develop simple equations as a part of the material.
31:38
Okay. I think the question is, do I have to know what ammonia is?
31:46
Okay. So I think I know what you mean. If I write in words the chemicals, do I have to know what they are? The answer is yes, I will only write down the chemicals that you should know. And not some kind of exotic thing that you've never heard of. It should be ridiculous. Okay. Another one. This is
32:10
the same kind of calculation with a slightly different content here, so again, I'm going to write it in words. Again, don't be flustered when things start to change. Let's have a look. This is
32:21
exactly the same question, slightly different formulation. I get in a reaction, in this case, dinitium pentoxide crystals, and they react with water, and they form aqueous nitric acid. The question is, what is the liquidation and what is the mass of the product?
32:41
What is different in this particular question is one of the reactants, the mass is given, the other reactant is actually water, which is a salt, too. So water is the one that's reacting, and it's a salt at the same time. You have two liters of it, and there's no mass left.
33:01
But there is a density difference here. So from the density you can calculate the mass, and from the mass you calculate the moles, and the mass. So that's really what I'm talking about, too. But if things are slightly different from the moles, then don't be confused. This is the same kind of question. All the information is there.
33:21
Okay. First things first, I'd like you to see what is the chemical reaction here. Dinitium pentoxide, that is a compound you should be able to write down. There it is. Dinitium pentoxide, it's five oxygens and two nitrogens, plus water, and look at the physical states I have to put there.
33:44
Solid or liquid. And this is my product. And that is dissolved in the AP space. Okay. Now to be sure that this one is balanced, let's quickly check if that is true. I have twice the six
34:01
oxygens, I have one here, five there, and six, two nitrogens, two nitrogens, okay, this one is balanced. So I have five. Second, I have to determine the elementary engagement, and I start by looking at the equation itself, and look at the mole ratio. So I see that from each one mole of this guy, I have one mole of water that
34:25
I need to completely react. So the reaction is one to one because the stoichiometric coefficients are one. All right. So the next method is to prepare that with the actual mole ratio. I have the mass here, I can divide this mass
34:43
by the molar mass of that hydrogen dioxide, and I use this information here, the liters and the density to calculate the number of moles of water. Now, intuitively, intuitively, you already know which one is going to be related. Would it be water, or would it be the other guy?
35:06
Two liters, that's a lot. It's one mole. So most likely is the dimension of the dioxide that is the mole that is going on first. But you have to double check. I always have to look at the question and try to understand
35:22
what is the scope here? What is the probably the right answer. Okay. So I'm going to take this, this is 40 grams, and I'm going to divide that by the molar mass. So converting kilograms to grams. So I like to have kilograms.
35:41
Thousand grams over kilograms. Then, divide by the molar mass, 108 for that hydrogen dioxide, and I find the number of moles, which happens to be 5.0. Next step, let's do the same for water. There it is.
36:03
Two liters of water, I have to go to a mass. Unfortunately, the density is given here. In grams per milliliter. So what I'm going to do first, I'm going to convert this to milliliters, it's a thousand milliliters in one liter, that means I have 2,000 milliliters. Two liters equals 2,000 milliliters.
36:25
And then I can multiply it by the density here. One gram per milliliter. There we go. One gram for one milliliter. So that means now I have my number of water molecules expressed in grams, and then I can multiply it by the density here.
36:41
In this case, in order to get the moles, I have to divide by the molar mass. And the molar mass of water is 18 grams per mole. So this calculation here should give me the number of moles of water. 110 moles of H2O. By the way, only once have I configured it, but I'm going to assume, for all kinds of purposes, that this is infinitely precise.
37:09
It's not going to be nicely phrased in this question. It's not the limited step in the calculation. I'm going to assume it.
37:22
Okay. So I have to kind of prepare these two guys, look at the ratio, there it is. Reagent number one, reagent number two, the ratio equals 0.045, that is much, much less than one, that means the numerator is the limiting reagent. That is this guy. And my prediction was correct. I expected that water was not the limiting reagent. It was the other guy.
37:47
Okay. So I take the number of moles of this to determine how much product I afford. I'm going to start my calculation by writing down the number of moles of dinitrogen peroxide. I have 5 moles, I want to convert that into the number of moles of the product.
38:14
For each one mole of dinitrogen pentoxide, I generate two moles of nitric acid. Nitric acid goes on top, this guy goes at the bottom.
38:24
So this will strike out, and effectively I multiply by two. Two moles for each one mole of this. And that's a total of 10 moles of dinitrogen pentoxide, 10 moles times the molar mass of sorry, this is nitric acid,
38:42
times the molar mass of nitric acid gives you the amount of grams of nitric acid. Same steps as we took before. From moles of one compound to moles of another compound, through the mole ratio, multiplying by the molar mass of your final compound. In this case, the answer is 630 grams of nitric acid.
39:07
So in this case, that's the end of this question. Okay. So recognize the flow in these kinds of calculations. You do the same thing over and over again, be sure you can make these conversions. All right. Last topic is this. Let's go back to one of our favorite compounds, which
39:32
is this fellow over here, this beautiful orange ammonium dichromate. Let's assume that I have one mole left to get it.
39:45
for my compound. And let's say I give you this compound, and I say, let's synthesize this kind of chromium peroxide. And you go to the lab, you do your best, you put your gloves on, you work the whole day, and at the
40:04
end of the day, you say, look, this is what I get. I get my beautiful chromium peroxide, I get 48 moles. All right? And I say, hold on a second. What's going on? I gave you one mole of this, and I'm going to
40:21
give you 48 moles of that. One mole of this should be given one mole of that. The ratio should be one to one. Well, you say, well, of course it's all good and well, but I can't even interpret it.
40:42
So, in reality, this is what I really get. One is a theoretical yield, but the actual yield is less than what I theoretically get. And that is very, very real life situation. This happens in the lab all the time. So, the percent yield is a quantifier for that.
41:04
It says the actual yield divided by the theoretical yield times one mole. So in this particular case, the actual yield was 0.8 moles, and the theoretical yield said you should have one, and if you multiply that by one percent, you get 80%. So, in this case, you did a pretty good job. You got a percent yield of 80%, which
41:25
is very decent for an aspiring case. So we have to be able to do a couple of calculations with percent yield. This is all there is to it, really. This is the definition of percent yield.
41:43
So let's do one quick calculation to implement it. Here it is. It's a reaction, of course, calcium hydride is going to react with water. Look, in excess, that's correct. The question already tells me which
42:01
one is in excess and which one is going to react. I'm going to determine that in this particular case. And the product is calcium hydroxide precipitate, and the other product is hydrogen gas. It also says, look, the yield is 84.3 Now given this yield, what is the mass of calcium hydroxide that you synthesize? All right. So let's do this.
42:25
First, this is always the first step. I want to know what is the reaction that we are talking about. Calcium hydride is this compound right here. Calcium and two hydrogen atoms. Calcium 2+, and the hydride is H-, two of those will give you a neutral compound.
42:44
It's solid because that's what the question says, so don't forget the state of the material. Water is liquid, so I put an L there, and this is the product, calcium precipitate, calcium hydroxide, and this is hydrogen gas. The reaction is already balanced. You can quickly see that. The number of hydrogens is balanced on both sides.
43:06
Okay. So let's assume that the yield is 100%. How much would I generate? Well, if you want to calculate how much you generate, you have to first know the length of the agent, and then convert the length of the agent into moles of the agent, and then into moles of the product, and then from moles of the product into grams of the product. Right? So let's do that.
43:28
Limiting the agent is given, it is calcium hydride, I have in the beginning 3.74 grams, I have to convert that as before, in moles. Okay? By dividing by the molar mass. 42.1 is the molar mass of calcium hydride.
43:51
Okay. Next, I want to see how much product I've formed by looking at the mole ratio of the reactant with the product that is one-to-one. Each one mole of calcium hydride will generate one mole of calcium hydroxide.
44:08
Okay? So this whole thing here is now the number of moles of calcium hydroxide. And I can calculate the number of calcium hydroxide, the total of calcium
44:21
hydroxide excess in grams. And multiplying by the molar mass. So it will be the same conversion as it was. A total of 6.58 grams of calcium hydroxide. So what is the next step I have to do? Let's have a quick look again at the question.
44:42
The question says what is the mass of calcium hydroxide that you would form in this case, given that you have a yield of 84%? This is what I have, what I would have gotten if I had 100%. I don't have 100%. I have less. I have 84%. So what I have to do, really, is to take 84% of this.
45:05
So the actual yield is 84%, expressed by its fraction, which is 0.843 times the mass, which this is basically the theoretical yield this is the theoretical yield, this is the actual yield, and this is the percentage expressed by
45:24
the fraction. So I rearrange the formula of percent yield. Theoretically, this is the reaction of the fraction. This is the fraction yield, and this is the actual number that you get in terms of grams.
45:43
So if you multiply these two numbers, you find 5.55, and that is the actual amount that you get.