If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Worked example: Calculating the amount of product formed from a limiting reactant

AP.Chem:
SPQ‑4 (EU)
,
SPQ‑4.A (LO)
,
SPQ‑4.A.1 (EK)
,
SPQ‑4.A.2 (EK)

## Video transcript

- [Instructor] So right here, we have a reaction where you can take some carbon monoxide gas and some hydrogen gas, and when they react, you're gonna produce methanol, and this is actually pretty interesting. Methanol has many applications. One of them it's actually race car fuel. But we're gonna do is study how much methanol we can produce if we have a certain amount of carbon monoxide and molecular hydrogen. So let's say, we have 356 grams of carbon monoxide and 65.0 grams of molecular hydrogen. Pause this video, and based on this, figure out how many grams of methanol will we produce? Well, a good place to start is by converting these numbers of carbon monoxide, this amount of carbon monoxide, and molecular hydrogen into moles. And so to do that, we can take out a periodic table of elements and the molar mass of carbon monoxide, you can look at the molar masses of carbon and oxygen and add them together. So 12.01 plus 16, that is going to be 28.01 grams per mole. And if we want to convert to moles, we're gonna have to multiply this times moles per gram. And so for every one mole, we have, we just figured it out 28.01 grams. And this is going to be approximately equal to 356 divided by 28.01 is equal to, let's see, and we have three significant figures here and four here. So I'll round to 12.7 moles, approximately 12.7 moles. And then we could do the same thing for the molecular hydrogen. And here we're going, for every one mole, how many grams or what's our molar mass of our molecular hydrogen? Well, each hydrogen atom is 1.008 grams per mole, but each molecule of hydrogen has two hydrogens in it, so it's gonna be two times six, so 2.016. 2.016 grams per mole or one mole for every 2.016 grams. And so this is going to be approximately equal to, get the calculator out again, 65 divided by 2.016 is equal is to that. And we have three significant figures, four significant figures, so if I round to three, it's approximately 32.2 moles, so 32.2 moles. And so the first thing to think about is, in our reaction for every one mole of carbon monoxide, we use two moles of molecular hydrogen, and then that produces one mole of methanol right over here. And so however much carbon monoxide we have in terms of moles, we need twice as much hydrogen. And so we see here a molecular hydrogen. And so two times 12.7 is going to be 25.4. So we actually have more than enough molecular hydrogen. And so we are going to use 25.4 moles of molecular hydrogen. How did I do that? Well, it's gonna be twice the number of moles of carbon monoxide, twice this number right over here is this right over here. And we can immediately see how much we're gonna have leftover. We're going to have leftover, leftover 32.2 minus 25.4 is 6.8, 6.8 moles of molecular hydrogen. And how many moles of methanol we're gonna produce? Well, the same number of moles of carbon monoxide that we're using up. It's a one-to-one ratio. So we're going to produce 12.7 moles of methanol. And so let me write that a year. So if I have 12.7 mole of methanol, CH3OH, how do I convert this to grams? Well, I have to multiply this times a certain number of grams per mole so that we can cancel out the moles or essentially the molar mass of methanol, and to figure out the molar mass of methanol, we'll get our calculator out again. So we have four hydrogens here. So four times 1.008 is going to be that. And then to that, we're going to add the molar mass of carbon 'cause we have one carbon plus 12.01, and then plus one oxygen in that methanol molecule is equal to that. And let's see, we will round to the hundreds place because our oxygen and carbon molar mass is only went to the hundreds place here, so 32.04, 32.04 grams per mole. So we have 12.7 moles times 32.04 grams per mole will tell us that we are going to produce that much methanol. And let's see, we have three significant figures, four, so I'll round to three. So approximately 407 grams of methanol, 407 grams of CH3OH. Now the next question is, what's the mass of hydrogen that we have leftover? Well, we just have to convert our moles of hydrogen that we have leftover to grams, 6.8 moles of molecular hydrogen times, the molar mass here in grams per mole is just gonna be the reciprocal of this right over here, so times 2.016, 2.016 is going to give us this right over here. And if we were rounding to two significant figures, which I have right over here, that is going to give us approximately 14 grams, approximately 14 grams of molecular hydrogen is leftover, leftover. So we used a good bit of it. We used about 51 grams and we have 14 grams leftover and it was carbon monoxide that was actually the limiting reactant here.
AP® is a registered trademark of the College Board, which has not reviewed this resource.