Stoichiometry: Mass-to-mass and limiting reagent

0:00- Let's solve a cool stoichiometry problem. Consider the chemical reaction where we have propane that burns with oxygen, giving us water and carbon dioxide. Now, if you're conducting this reaction, our question is how much propane in grams is needed to completely combust with 256 grams of oxygen in the given reaction?

0:20The first step to doing any stoichiometry problem is to ensure that we have a balanced chemical reaction. And you can check over here that this reaction is actually balanced, all right? 'Cause once we have that, then we can write the mole ratio. We now know that one mole of propane will always react with five moles of oxygen, giving me four mos of water and three moles of carbon dioxide.

0:39For this particular reaction to work, this mole ratio is necessary. This means that propane and oxygen will always be consumed, regardless of whatever quantity they have, they will always be consumed in a one to five ratio. That's what it means, okay? Now let's look at what's asked.

0:54We need to calculate how much propane is necessary to completely combust 256 grams of oxygen. So we have 256 grams of oxygen. We want to completely use it up. We wanna completely combust it. So for that, how much propane do we need? Now the first thing that comes to my mind is I'm thinking, wait a second, wait a second.

1:14This is pretty straightforward, right? I mean, I know that the ratio in which propane and oxygen must be is one to five, meaning propane must be one fifth of oxygen. So the same thing applies over here. The amount of propane necessary must be one fifth of how much oxygen we have.

1:27So it should be one fifth of this, right? Right? Well, that would be a mistake because this is the ratio of moles. This only works when you're dealing with moles. But notice we are given in terms of grams, we're given the mass of the oxygen, not moles of oxygen over here.

1:43Remember that a mole of oxygen would have a different mass compared to a mole of propane. So the ratio for the mass would be different. So I can't just say that this will also be one to five ratio and solve it this way. So what do we do? Hey, if I could somehow convert this amount of oxygen that is given to me in terms of moles, if I can figure out how many moles of oxygen that is, then I can use the conversion factor over here to convert from moles of oxygen to moles of propane.

2:15And this is something that we've seen in our previous video. We've seen how to convert from moles of one participant of the reaction to the moles of another participant, right? We have solved a couple of problems on this, okay? So once we do that, now finally, since I want grams of propane, I can convert back from moles of propane to grams of propane and I'll be done.

2:36So this is what you're gonna do. So the big question now is how do I do these conversions? How do I go from grams to moles and moles to grams? Hey, we know how to do that as well by using the idea of molar mass. Again, in our earlier videos, we've seen how to calculate molar masses of things, like of molecules, and we're not gonna do that over here.

2:54We're just gonna look at a table which gives us the molar mass. So the molar mass of propane is 44.1 grams per mole, meaning one mole of propane has a mass of 44.1 grams. Same as the case with oxygen. So now the strategy is we're gonna use this as our conversion factor to go from here to here, and then we use this as a conversion factor to go from here to here like we did our previous video.

3:17And again, we're gonna use this as a conversion factor to go from here to here. Alright? So it'll be a great idea to pause the video and see if you can try this yourself. It's fine if it looks daunting for now, and don't worry about getting the right answer. Just attempt it and then we'll solve it together.

3:33All right. Here it goes. So we start with what's given to us. You are given 256 grams of oxygen. And now my first step is to convert from grams to moles. And I'm gonna use this as a conversion factor, okay? And the way I'm gonna think of, the way I'm gonna write this is I need moles, so I need moss on the numerator.

3:49I wanna convert this to moles, so moles comes on the numerator. I wanna get rid of the grams. So my conversion factor should have grams in the denominator. So I will write this as moles per gram. So it's gonna be one mole per 32 grams. That's how I write it, 'cause when I write it this way, look, the grams cancel out and I'm left with moles of oxygen.

4:09So this number now represents the moles of oxygen. And just to show you the power of dimensional analysis, okay, imagine we got this wrong, imagine we got confused and we wrote it as a reciprocal. So imagine we wrote it not like this, but we wrote it as 32 grams of oxygen per mole of oxygen, okay?

4:28We just wrote as it is. Now, immediately when I am looking at the units, I realize, hey, wait a sec, wait a second, grams is not canceling out. Oh, no, no, no, grams should have been in the denominator. Oh, the moles should have been in the numerator. And that's how I understand that, no, this should be the reciprocal.

4:44So if I write out the entire units, not just grams, but grams of O2, the moles of O2, the whole unit, then I can use my dimensional analysis where I can just look at my units and then I can see whether I'm on the right track or not. Anyways, now this number, whatever this number is, it represents the moles of oxygen that I have.

5:04And I can plug it in my calculator and figure out how much this number is. But let's not do that because we are not done with our calculation. Let's write out the whole thing and then let's input everything in our calculator together. So we are here now, we have found the number of moles.

5:20This number represents the number of moles of oxygen. The next step is to convert from moles of oxygen to moles of propane. How do I do that? Well, this time we use this as a conversion factor, okay? And again, I'm gonna write it in such a way that I need moles of propane in the numerator because that's what I want.

5:36And I want to now get rid of moles of oxygen, so the moles of oxygen comes in the denominator. So I'm gonna write it as one mole of propane per five moles of oxygen. This way, again, by looking at the units, I can say, Hey, the moles of oxygen cancels out and now I'm left with moles of propane.

5:54So this now number represents the amount of propane I need for complete, the moles of propane I need for the complete combustion. And the final step is I need this in grams. So I need to convert this in grams. So I'm gonna use this as my conversion factor. And again, since I want grams, I want grams on the numerator, and I wanna get rid of the moles, I want moles in the denominator so I can cancel it out.

6:17So I'll write this as 44 by one grams per one mole of propane. So this way the moles cancel out and I am now left with grams of propane.

6:31- [Instructor] And so finally, we can plug this number into our calculator. So it's 256 times one times one times 44.1, divided by 32.0, divided by five, divided by one. So there you have it, 70.56. But we've got to round it off to three significant figures, because we go to our data, that's the least amount of significant figures our data has.

6:58Three, right? So we gotta round this off to three significant figures, giving me 70.6. And so there we have it.

7:06- [Instructor] So 70.6 grams of propane is necessary to completely combust with 256 grams of oxygen. And now before we jump to another problem, we can use these numbers to understand a cool concept called the limiting reactant. So let me hide this for a while.

7:23So imagine we had 85.3 grams of propane. We reacted that with 256 grams of oxygen. What would've happened? Well, we already know that we need only 70.6 grams of propane to completely combust with 256 grams of oxygen. So once this much amount of propane is used up, all of oxygen would've already used up, and then the reaction would stop, because there is no more oxygen to consume, which means there will be some propane left over, right?

7:52How much leftover would it be? Well, you just subtract these two and you'll get how much the propane would be left over, which means in this particular reaction, the propane would be in excess quantity. And since it's the oxygen that runs out fast, it's the oxygen that is in the limiting quantity.

8:08And therefore we say oxygen is the limiting reactant. Even if I had a thousand grams of propane, only two six grams of oxygen is there to react with 70.6 grams of propane. So only 70.6 grams out of that a thousand grams would react with 256 grams of oxygen, and the rest will just stay as it is.

8:26But what if instead of 85.3 grams, what if I had 60.2 grams of propane? What would've happened then? Ooh, then notice, I don't even have enough propane to combust with 256 grams of oxygen, which means this time propane is gonna run out first, and it would not consume the entire 256 grams.

8:47Some of the oxygen would remain because I need 70.6 grams to completely consume 256 grams of oxygen. Since I have less than that, less than 256 grams would be consumed. So now propane will be completely consumed, which means oxygen is in the excess quantity. Propane now becomes the limiting reactant.

9:06So if we were given masses of two reactants and we are asked to figure out which of them is the limiting reactant, how would we figure this out? Well, we can calculate how much propane is necessary to completely combust with 256 grams of oxygen, which means we do the same calculations, figure that number out, and then compare what's given.

9:29If we have less than what's necessary, then that itself becomes the limiting reactant. If there is more propane than what's necessary, then oxygen would be the limiting reactant. And once I know which of them is the limiting reactant, I know that that's the one that's gonna be completely consumed, and then I can use that to figure out how much, say, water is produced, or how much carbon dioxide is produced.

9:54I cannot use the one that's in excess because not all of it is going to be consumed. Okay, let's just do another problem on this idea and it'll make sense. This time, we are reacting nitrogen with hydrogen to get ammonia. And again, what's given to us, 12.2 grams of nitrogen reacts with 3.25 grams of hydrogen, and we have to calculate how many grams of ammonia can be theoretically reduced.

10:17All right? So again, the equation is balanced. That's the first step. Then we'll write down the mole ratio. One mole of N2 would react with three moles of H2 to give two moles of ammonia. And then I write down what is given to us. We are given that we have 12.2 grams of nitrogen reacting with 3.25 grams of hydrogen, and we need to figure out how much ammonia we get.

10:37The first step over here is to figure out which of this is the limiting reactant and which of them is in excess. Because it's only the limiting reactant that will be completely consumed, and I can use that to convert to figuring out how much ammonia we get, all right?

10:50So how do we figure out which is a limiting reactant over here? Well, again, in general, there are many ways to do it, but over here we are gonna calculate how much hydrogen we need, how many grams of hydrogen we need to completely react with nitrogen. And then we'll look at that number and compare with what's what we have.

11:07And then we'll decide which is in excess, and which limiting. So what we're gonna do, oops, okay, so what we're gonna do is we're gonna take the same thing that we did before. We'll first convert from the grams of nitrogen to moles of nitrogen using this, then moles of nitrogen to moles of hydrogen, and from moles of hydrogen to grams of hydrogen to figure out how much hydrogen we need to completely react with this.

11:29Again, feel free to pause the video and see if you can try and do it yourself. At any moment, you can pause, but let's do this now. So we start with what is given to us, 12.2 grams of nitrogen. To convert it into moles of nitrogen, I'm gonna use this, I need moles on the numerator, and I need to cancel out the grams in the denominator.

11:45So I'm gonna write this as one mole of nitrogen per 28 grams of nitrogen. And so now I can cancel out the grams, dimensional analysis. Now I have moles of nitrogen. I want to convert from moles of nitrogen to moles of hydrogen. To do that, I'm gonna use this as my conversion factor.

12:02And since I want moles of hydrogen on the numerator, I write this as three moles of hydrogen per one mole of nitrogen. So that way, the moles of nitrogen cancels out and now I have moles of hydrogen. And finally, to go from moles of hydrogen to grams of hydrogen, I'll use this as my conversion factor.

12:18I need grams on the numerator because I want grams and I want to cancel moles, so that should be in the denominator. So I'll write this as grams per mole. So 2.02 grams per one mole of hydrogen. The moles cancel out, and boom, this number now tells me how many grams of hydrogen is necessary to completely react with 12.2 grams of nitrogen.

12:38So just plug this number in the calculator, and if we do that, we get 2.64 grams of hydrogen. That's how much is necessary to completely consume nitrogen, to completely react to nitrogen. So now look at how much hydrogen we have. We have way more than what is necessary, which means the hydrogen is in excess, and that means nitrogen must be the limiting reactant.

13:06This means the entire 12.2 grams of nitrogen will be completely consumed in this reaction, and now I can use that to figure out how much ammonia I'll get. So I have to convert from nitrogen to ammonia, same steps, but I'm gonna use the ratio of nitrogen to ammonia for that.

13:21So again, to start with, convert from nitrogen moles to grams, the same thing that I did over here. So I'll multiply it by one mole of nitrogen, divided by 28.0 grams of nitrogen. That cancels out the grams, and I'm now left with moles. The next step is to convert from moles of nitrogen to moles of ammonia.

13:40That's what I want. So moles of ammonia, for that, I'm gonna use this as the conversion factor. And since I want moles of ammonia, I'll use, this should be in the numerator. So I'll write two moles of ammonia divided by one mole of nitrogen. Here we go. That cancels out the moles of nitrogen.

13:56And finally, to go from moles of ammonia to grams of ammonia, I'm gonna use this, the final, final conversion factor. Okay, so for that, I want grams on the numerator and moles on the denominator. So write 17 grams per mole. For ammonia, the moles cancel out, and there we have it.

14:17This will now tell me the amount of, the grams of mole that is produced in this reaction. So if I plug this in my calculator, I end up with 14.8 grams of ammonia. This is my final result. And again, remember, this is the theoretical yield. On paper, this is how much we should get if we completely consume 12.2 grams of nitrogen.

14:39But the actual yield you might get in your lab when you do this experiment could be less than this. 'Cause we've seen before, the reasons could be incomplete reaction, not all the nitrogen might be consumed, and there could be some impurities over here as well.