Main content

## Chemistry library

### Course: Chemistry library > Unit 5

Lesson 3: Limiting reagent stoichiometry- Limiting reactant and reaction yields
- Worked example: Calculating the amount of product formed from a limiting reactant
- Introduction to gravimetric analysis: Volatilization gravimetry
- Gravimetric analysis and precipitation gravimetry
- 2015 AP Chemistry free response 2a (part 1 of 2)
- 2015 AP Chemistry free response 2a (part 2/2) and b
- Limiting reagent stoichiometry

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# 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)

In a chemical reaction, the reactant that is consumed first and limits how much product can be formed is called the limiting reactant (or limiting reagent). In this video, we'll determine the limiting reactant for a given reaction and use this information to calculate the theoretical yield of product. Created by Sal Khan.

## Want to join the conversation?

- What does the "(g)" and the "(l)" stand for?(3 votes)
- They stand for the physical states. (s) is for solid, (l) is for liquid, (g) is gas, and (aq) is aqueous. Hope that helps.(29 votes)

- I don't understand why you would multiply the CO by two (starting around minute2:45of the video). I understand that you need 1 mole of CO for every 2 moles of H2. But it seems like you should then multiply the H2, not the CO, by 2 instead.(2 votes)
- The previous replier is correct, just want to extra some extra bits. Of the two reactants, the limiting reactant is going to be the reactant that will be used up entirely with none leftover. For the CO if you were to use it up completely you would use up 12.7 mols of CO. You need twice as much H2 as CO since their stoichiometric ratio is 1:2. So 12.7 mols of CO would require twice the amount of mols of H2, or 25.4 mols of H2. If you wanted to use up the entire 32.2 mol supply of H2, you would need 1/2 of the 32.2 mols for the required mols of CO, or 16.1 mols of CO. But since we don't have 16.1 mols of CO, we have less than that, the CO is therefore the limiting reactant. So we multiply the CO mols by two to know how many moles of H2 gas would be required to react with the entirety of the CO supply because of that 1:2 ratio. If we had done it the other way and multiplied the 32.2 mols of H2 by 2 instead, that would tell us we need 64.4 mols of CO to react which doesn't make sense since we know we need more moles of H2 than mols of CO for the reaction to occur. Hope that helps.(17 votes)

- Why are we going to produce 12.7 mole of methanol?(3 votes)
- Since carbon monoxide is our limiting reactant then the theoretical yield is going to be determined by how many moles of carbon monoxide we have. Again the amount of product produced, in this case methanol, is limited by the amount of limiting reactant we have since the reaction cannot proceed if we have no more carbon monoxide. So if we have 12.7 moles of carbon monoxide and carbon monoxide, and methanol have a 1:1 mole-to-mole ratio in the balanced chemical equation, then we will also produce 12.7 moles of methanol as our theoretical yield.

Hope that helps.(6 votes)

- After you've turned the grams of the reactants into moles of reactants and have found the limiting reactant, you would multiply by the mole-to-mole ratio. It's part of dimensional analysis which lets you do successive conversions like this by either multiplying or dividing. You cross cancel units like you would in fractions to get the units you desire.

Using your example, let's say they gave us grams of a reactant (and we assume we already know it's the limiting reactant) and want us to find the theoretical yield of a product. Where the reactant has a coefficient of 4 and the product has one of 2. The idea would be to turn grams of the reactant into moles of the reactant using the molar mass of the reactant, then multiply by the mole-to-mole ratio in the balanced chemical equation to get moles of your product. And then we could also find the grams of the product if they asked us to by using the molar mass of the product. The math would look as follows: (reactant grams/1) x (1 mol reactant/ reactant grams) x (2 mol product/4 mol reactant) x (product grams/1 mol product). So if you follow it you can see each unit diagonal to the same of its kind is cross canceled each time we multiply or divide to get a new desired unit. It also gets rid of any doubt whether we have to divide or multiply in a step. So we see that if we divide our original grams of reactant by the molar mass, we get moles of our reactant. Then multiply those grams by 2:4 which is the ratio of products to reactants to get moles of product. Finally we multiply the moles of the product by the molar mass to get the grams of our product.

I understand it but don't(3 votes) - hello, I am confused about how you would find the theoretical yield if the ratio wasn't one to one for the limiting reagent. But say 4 on the reagent side and 2 on the product side?(1 vote)
- After you've turned the grams of the reactants into moles of reactants and have found the limiting reactant, you would multiply by the mole-to-mole ratio. It's part of dimensional analysis which lets you do successive conversions like this by either multiplying or dividing. You cross cancel units like you would in fractions to get the units you desire.

Using your example, let's say they gave us grams of a reactant (and we assume we already know it's the limiting reactant) and want us to find the theoretical yield of a product. Where the reactant has a coefficient of 4 and the product has one of 2. The idea would be to turn grams of the reactant into moles of the reactant using the molar mass of the reactant, then multiply by the mole-to-mole ratio in the balanced chemical equation to get moles of your product. And then we could also find the grams of the product if they asked us to by using the molar mass of the product. The math would look as follows: (reactant grams/1) x (1 mol reactant/ reactant grams) x (2 mol product/4 mol reactant) x (product grams/1 mol product). So if you follow it you can see each unit diagonal to the same of its kind is cross canceled each time we multiply or divide to get a new desired unit. It also gets rid of any doubt whether we have to divide or multiply in a step. So we see that if we divide our original grams of reactant by the molar mass, we get moles of our reactant. Then multiply those grams by 2:4 which is the ratio of products to reactants to get moles of product. Finally we multiply the moles of the product by the molar mass to get the grams of our product.

This works for any situation where you have to convert multiple times. Specifically here we just have to add that ratio of products to reactants step in the middle of the conversions if the ratio wasn't 1:1. Hope that helps.(6 votes)

- Hello - Is it possible to work backwards - If we have the amount of products in grams can we figure out how much of the reactents we need as well as the limiting agent?(2 votes)
- Yep, you can do exactly that. The idea being usually that you have a desired amount of product you wish to produce from a reaction and you'd like to predict how much of the reactants it will take to do so. And then you can compare how much of the reactants you'd need to how much of the reactants you actually have to determine the limiting reactant.

Hope that helps.(4 votes)

- Aren't we supposed to not care about significant figures until the end of calculation since otherwise that would bring in even more error, contrary to4:50? Or you have to do that when switching between multiplication/division and addition/subtraction?(2 votes)
- Yes, you're correct. Technically we should only round off our numbers at the very last calculation or else we could invite error into our final answer. Just by coincidence rounding off early here doesn't change the final answer. But we can imagine that if we had more precise measurements which allowed us to use more sig figs for our answer, like four, then yes we would see an error in our answer if we rounded prematurely.

Hope that helps.(3 votes)

- How did you get the leftover H2? Did you somehow calculate it from 12.7?(1 vote)
- So with CO as the limiting reactant, we know we'll use the entirety of the 12.7 mols of CO. Since CO and H2 is a 2:1 mol ratio from the balanced equation, we know we'll have to use twice as much H2 as we do CO, or 25.4 mols of H2. If we had 32.2 mols of H2 initially to work with, then the leftover H2 is the differernce between the mols of H2 we initially had and the mols used up in the reaction, or 32.2 - 25.4 = 6.8 mols of excess H2. Hope that helps.(4 votes)

- What if only one of the elements reacting to each other has a given mass?(1 vote)
- I believe you mean chemical species instead of elements because we can have a compound with multiple elements in them which we are given mass information about. For this reaction we have molecular hydrogen which is different from elemental hydrogen, and we have carbon monoxide which is a compound formed from two elements; carbon and oxygen.

But I get what you're asking, what if only know the mass of a single reactant and not both? Which one would be the limiting reactant then if we don't know how much of one reactant we have? Usually if only a single reactant's mass is given it is because they specify that that other reactant is added in excess or we have an unlimited supply of it. If we use this reaction in the video as an example, lets say they give 65.0 grams of hydrogen gas but say that we have an excess of carbon monoxide. It means that we have more carbon monoxide than we would ever need for the reaction to progress. This would mean that the hydrogen gas is the limiting reactant and would dictate the theoretical yield because we essentially have an unlimited supply of carbon monoxide.

The other case would be if they give the volume of a gaseous reactant instead of the grams from which we can calculate the moles of the gas using the ideal gas law.

Hope that helps.(3 votes)

- what if there are 2 products how would you do that(1 vote)
- same thing... once you find out what the limiting reagent is, as long as you know the relative number of moles (which is specified by the big number before a chemical formula), you can find out the theoretical yield by multiplying the number of moles by the Relative atomic mass (Mr) of the product you want to find out(2 votes)

## 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.