Worked example: Calculating the amount of product formed from a limiting reactant | Khan Academy

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 going to produce methanol. This is actually pretty interesting; methanol has many applications. One of them, it's actually race car fuel. But what we're going to 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 oxide, and molecular hydrogen into moles.

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: 12.01 plus 16, that is going to be 28.01 grams per mole. If we want to convert to moles, we're going to have to multiply this times moles per gram.

For every one mole, we have—we just figured it out—28.01 grams. This is going to be approximately equal to 356 divided by 28.01. 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.

Then we could do the same thing for the molecular hydrogen. 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 going to be two times this, so 2.016 grams per mole or one mole for every 2.016 grams.

This is going to be approximately equal to—get the calculator out again—65 divided by 2.016. That is equal 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.

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. However much carbon monoxide we have, in terms of moles, we need twice as much hydrogen. We see here, molecular hydrogen, and so 2 times 12.7 is going to be 25.4.

So we actually have more than enough molecular hydrogen. We are going to use 25.4 moles of molecular hydrogen. How did I do that? Well, it's going to be twice the number of moles of carbon monoxide, twice this number right over here—this right over here.

We can immediately see how much we're going to have left over: we're going to have left over 32.2 minus 25.4, which is 6.8 moles of molecular hydrogen. And how many moles of methanol are we going to 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.

Let me write that here. If I have 12.7 moles of methanol (CH₃OH), how do I convert this to grams? We'll 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.

To figure out the molar mass of methanol, we'll get our calculator out again. So we have four hydrogens here, so 4 times 1.008 is going to be that. Then to that, we're going to add the molar mass of carbon because we have one carbon (plus 12.01) and then plus one oxygen in that methanol molecule (is equal to that).

Let's see; we will round to the hundredths place because our oxygen and carbon molar masses only went to the hundredths place here, so it's 32.04 grams per mole. We have 12.7 moles times 32.04 grams per mole, which will tell us that we are going to produce that much methanol.

Let's say 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 left over?

We just have to convert our moles of hydrogen that we have left over to grams. 6.8 moles of molecular hydrogen times the molar mass you're in grams per mole is just going to 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 left over. So, we used a good bit of it; we used about 51 grams and we have 14 grams left over. It was carbon monoxide that was actually the limiting reactant here.