Stoichiometry: mass-to-mass and limiting reagent | Chemistry | Khan Academy

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. Okay, now if you're conducting this reaction, our question is how much propane in grams is needed to completely combust with 256 g of oxygen in the given reaction?

Now, the 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, because once we have that, then we can write their mole ratio. We now know that one mole of propane will always react with five moles of oxygen, giving me four moles of water and three moles of carbon dioxide. For 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, but they will always be consumed in a 1:5 ratio. That's what it means.

Okay, now let's look at what's asked. We need to calculate how much propane is necessary to completely combust 256 g of oxygen. Okay, so we have 256 g of oxygen, we want to completely use it up, we want to completely combust it. So for that, how much propane do we need? Now, the first thing that comes to my mind is I'm like thinking, "Wait a second, wait a second, this is pretty straightforward, right?" I mean, I know that the ratio in which propane and oxygen must be is 1:5, meaning propane must be 1/5 of oxygen. So the same thing applies over here; the amount of propane necessary must be 1/5 of how much oxygen we have. So it should be 1/5 of this, right?

Well, that would be a mistake because this is the ratio of moles. This only works when we're dealing with moles. But notice we are given in terms of grams; we are given the mass of the oxygen, not moles of oxygen over here. Remember 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 1:5 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 could figure out how many moles of oxygen there are, then I can use the conversion factor over here to convert from moles of oxygen to moles of propane. This is something that we've seen in our previous video; we've seen how to convert from moles of one participant in 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. Okay, so this is what you're going to 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 molecules, and we're not going to do that over here; we're just going to look at a table that gives us the molar mass.

So the molar mass of propane is 44.1 g per mole, meaning one mole of propane has a mass of 44.1 g. The same is the case for oxygen. Okay, so now the strategy is we're going to 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 in our previous video, and again we're going to use this as a conversion factor to go from here to here.

All right, 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. All right, here goes.

So we start with what's given to us. We are given 256 g of oxygen, and now my first step is to convert from grams to moles, and I'm going to use this as a conversion factor, okay? And the way I'm going to think of it is I need moles, so I need moles in the numerator. I want to convert these to moles, so moles come on the numerator; I want to 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 going to be 1 mole per 32 g. That's how I write it because when I write it this way, look, the grams cancel out and I'm left with moles of oxygen.

So 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 g of oxygen per mole of oxygen. Okay, we just wrote it as it is. Now the moment I look at the units, I realize, "Hey, wait, wait a second; grams is not canceling out." Oh no, no, no; grams should have been in the denominator. The moles should have been in the numerator, and that's how I understand that no, this should be the reciprocal.

Okay, so 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. Anyway, now this number, whatever this number is, represents the moles of oxygen that I have, okay? And 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.

Okay, so we are here now; we have found the number of moles. This 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 going to write it in such a way that I need moles of propane in the numerator because that's what I want, and I want to now get rid of moles of oxygen. So the moles of oxygen come in the denominator. So I'm going to 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. So this now number represents the amount of propane I need for complete combustion—the moles of propane I need for the complete combustion.

Okay, and the final step is I need this in grams, so I need to convert this in grams. So I'm going to use this as my conversion factor, and again, since I want grams, I want grams on the numerator, and I want to get rid of the moles; I want moles in the denominator so I can cancel it out. So I'll write this as 44.1 g per 1 mole of propane. So this way, the moles cancel out, and I am now left with grams of propane.

And so finally, we can plug this number into our calculator, so it's 256 * 1 * 1 * 44.1 divided by 32.0 divided by 5 divided by 1. So there you have it, 70.56. But we got to round it off to three significant figures because look at our data. That's the least amount of significant figures our data has—three, right? So we round this off to three significant figures, giving me 70.6. And so, there we have it. So 70.6 g of propane is necessary to completely combust with 256 g of oxygen.

Now, before we jump to another problem, we can use these numbers to understand a cool concept called the limiting reactant. Okay, so let me hide this for a while. So imagine we had 85.3 g of propane; we reacted that with 256 g of oxygen. What would have happened? Well, we already know that we need only 70.6 g of propane to completely combust with 256 g of oxygen. So once this much amount of propane is used up, all of the oxygen would have already been 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?

How much left over 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, propane would be in excess quantity. And since it's the oxygen that runs out—it's the oxygen that is in the limiting quantity—therefore we say oxygen is the limiting reactant. Even if I had 1000 g of propane, only 256 g of oxygen is there to react with 70.6 g of propane. So only 70.6 g out of that 1000 g would react with 256 g of oxygen, and the rest will just stay as it is.

But what if instead of 85.3 g, what if I had 60.2 g of propane? What would have happened then? Ooh, then notice I don't even have enough propane to combust with 256 g of oxygen, which means this time propane is going to run out first, and it would not consume the entire 256 g. Some of the oxygen would remain because I need 70.6 g to completely consume 256 g of oxygen. Since I have less than that, less than 70.6 g would be consumed.

So now propane will be completely consumed, which means oxygen is in the excess quantity, and propane now becomes the limiting reactant. So 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 g of oxygen, which means we do the same calculations, figure that number out, and then compare what's given. If 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.

Once I know which of them is the limiting reactant, I know that's the one that's going to 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. I 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 g of nitrogen reacts with 3.25 g of hydrogen, and we have to calculate how many grams of ammonia can be theoretically produced. All 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.

Then I write down what is given to us because we are given that we have 12.2 g of nitrogen reacting with 3.25 g of hydrogen, and we need to figure out how much ammonia we get. The first step over here is to figure out which of these 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, so how do we figure out which is the limiting reactant over here? Well, again, in general, there are many ways to do it, but over here we're going to 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 it with what we have, and then we'll decide which is in excess and which is limiting.

Okay, so what we're going to do—oops—okay, so what we're going to do is we're going to 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, figure out how much hydrogen we need to completely react with this.

Okay, again 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 g of nitrogen. To convert into moles of nitrogen, I'm going to use this. I need moles on the numerator, and I need to cancel out the grams in the denominator. So I'm going to write this as 1 mole of nitrogen per 28 g of nitrogen, and so now I can cancel out the grams—dimensional analysis.

Okay, now I have moles of nitrogen. I want to convert from moles of nitrogen to moles of hydrogen. To do that, I'm going to use this as my conversion factor, and since I want moles of hydrogen on the numerator, I'll write this as three moles of hydrogen per one mole of nitrogen. So that way, the moles of nitrogen cancel out, and now I have moles of hydrogen.

And finally, to go from moles of hydrogen to grams of hydrogen, I will use this as my conversion factor. I 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 2.22 g per 1 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 g of nitrogen.

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

Okay, this means the entire 12.2 g 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 going to use the ratio of nitrogen to ammonia for that. So again, to start with, convert from nitrogen moles to grams—the same thing that I did over here. So multiply it by 1 mole of nitrogen divide by 28.0 g of nitrogen; that cancels out the grams, and I have—I'm now left with moles.

The next step is to convert from moles of nitrogen to moles of ammonia—that's what I want. So moles of ammonia, for that I'm going to use this as the conversion factor. And since I want moles of ammonia, I want this to be in the numerator, so I'll write 2 moles of ammonia divided by one mole of nitrogen. Here we go, that cancels out the moles of nitrogen.

Finally, to go from moles of ammonia to grams of ammonia, I'm going to use this—my final, final conversion factor. Okay, so for that, I want grams on the numerator and moles on the denominator. So I'll write this as 17 g per mole for ammonia. The moles cancel out, and there we have it. This will now tell me the amount of grams of ammonia that is produced in this reaction.

So if I plug this in my calculator, I end up with 14.8 g 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 g of nitrogen. But the actual yield you might get in your lab when you do this experiment could be less than this because you'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.