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### Course: AP®︎/College Chemistry>Unit 4

Lesson 4: Stoichiometry

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

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?
• They stand for the physical states. (s) is for solid, (l) is for liquid, (g) is gas, and (aq) is aqueous. Hope that helps.
• I don't understand why you would multiply the CO by two (starting around minute of 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.
• 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.
• Why are we going to produce 12.7 mole of methanol?
• 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.
• How did you get the leftover H2? Did you somehow calculate it from 12.7?
• 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.
• 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
• 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.
• 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?
• 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.
• Why do you multiply the CO(12.7) mol by2 to get 25.4
• We need to know how much hydrogen gas, H2, is required to react with the entirety of the carbon monoxide, CO. Since CO is related to H2 in the chemical equation by a 1:2 ratio, the reaction consumes H2 at twice the rate it consumes CO. For every mole of CO, we need 2 moles of H2. So if we have 12.7 mole of CO then we need 2x12.7 or 25.4 mole of H2 to completely react with that CO.

Hope that helps.
• 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 to ? Or you have to do that when switching between multiplication/division and addition/subtraction?