Limiting Reagent: Study with Video Lessons, Practice Problems & Examples

In chemical reactions, the limiting reagent is the reactant that is completely consumed, determining the maximum theoretical yield of product, while the excess reagent remains after the reaction. To identify these, stoichiometry must be applied to each reactant, converting given amounts to moles and then to the desired product. This process allows for the calculation of both the limiting reagent and the theoretical yield, emphasizing the importance of understanding stoichiometric relationships in predicting reaction outcomes.

Limiting Reagent

In this series of videos, we're going to still talk about stoichiometry, but now we're going to introduce the idea of the limiting reagent. Now the limiting reagent, sometimes called the limiting reactant, is the reactant that is completely consumed in a reaction and determines the maximum amount of product. Now we start talking about limiting reactant or reagent when more than one reactant in our chemical equation is given a starting amount. Now, we're going to say here that the limiting reagent helps us to determine the theoretical yield. This is the maximum amount of product that can form from a chemical reaction. It's also referred to as the 100% yield or the maximum yield. Now, different from the limiting reagent is the excess reagent. Now this is the reactant that remains after the completion of the chemical reaction. So you have a reactant that acts as a limiting reagent, and then you can have another reactant acting as the excess reagent. Now in order to determine which reactant is which, you must work out the amounts of products each can make. And from there, we'll have enough information to tell which is which. Now that we've gotten the definitions, click on the next video and see how the stoichiometric chart changes slightly when dealing with limiting reagents and excess reagents.

Limiting Reagent

You might notice that this stoichiometric chart is slightly different from the one we learned about in stoichiometry. In stoichiometry, we're accustomed to getting only one given amount for a reactant within our chemical reaction. Sometimes we wouldn't even mention the other reactant. Now we're going to have chemical equations with reactants and more than one of them will have a given amount. So what we're going to have to do is we're going to have to do stoichiometry for each reactant. So for reactant 1, we're going to take its amount and that amount might be in grams, so we'll have grams given for it. Convert those grams given to moles given. We know that we'd have to do the jump to get some moles of unknown and to do that we have to use coefficients from the balanced equation. Those moles of unknown, we can convert them into ions, atoms, formula units, molecules, or grams. We get our answer for that amount of product or unknown, then we'd have to do it again for reactant 2. Go through the whole process and find out how much of our unknown we have. So from those amounts that we've calculated, we can determine which one of these reactants is the limiting reagent and which one is the excess reagent. We can also determine from the limiting reagent what the theoretical yield is. It's just double the work to help us find out how much product we're going to make. So, click on to the next video, and let's take a look at an example question where we put to practice this new idea of our stoichiometric chart.

Limiting Reagent Example 1

In this example question, it says, chromium(III) oxide reacts with hydrogen sulfide, which is H₂S gas, to form chromium(III) sulfide and water. Alright. So here we have our chromium(III) oxide reacting with 3 H₂S to produce chromium(III) sulfide and 3 H₂O. Within this question, they ask, what is the mass of chromium(III) sulfide formed when 14.20 grams of chromium(III) oxide reacts with 12.80 grams of hydrogen sulfide? Now in this question, they're giving us more than one gram of given. So we have 2 grams of given, and they're asking me to find grams of unknown. Because they're giving me multiple grams of given, I'm going to have to do stoichiometry for both of those amounts. So I'm going to go through the whole process of the stoichiometric chart with a 14.20 grams and then again with the 12.80 grams. Once we get our answers for the amount of chromium(III) sulfide produced, we'll talk about those two answers. And which one is the final solution? So if we take a look here at these steps, we have step 1, we're going to convert the given quantities, so both those grams, into moles of given. Now if any compound or compounds is said to be in excess, then just ignore. Once we have both of those moles of given, we do a mole to mole comparison to convert them into moles of unknown. And then step 3, we convert those moles of unknown finally into our desired units which is grams of chromium(III) sulfide. Alright. So let's set this up. We have 14.20 grams of chromium(III) oxide. Let's just start out with this one. I'm gonna convert moles of it, grams of it into moles. So 1 mole of chromium(III) oxide on top, its grams on the bottom. Here I need to know how much this weighs. So I have to look at the periodic table. So we have 2 chromiums and 3 oxygens. According to the periodic table, the atomic mass of chromium is 51.996 grams, and for oxygen, it's 16 grams. This comes out to 103.992 grams, and then this is 48 grams. So together that's 151.992 grams as the mass for chromium(III) oxide. So here, grams cancel out. Now I have moles of given. We're going to convert those moles of given into moles of unknown. So moles of our unknown is what we're asked to find which is chromium(III) sulfide. Remember at this point, this is where we'd utilize the jump. So we look at the coefficients of the balance equation. So it's for every one of this, there's one of this. So these moles cancel out and then finally we're going to convert 1 mole of chromium(III) sulfide into grams of chromium(III) sulfide. We already calculated the mass for chromium(III) oxide so let's do it for chromium(III) sulfide. So here we're gonna have, 3 sulfurs. Sulfur on the periodic table is 32.07 grams roughly. So multiplying that by 3 gives me 96.21. So bring everything down here, so that'd be 202.2 grams of chromium(III) sulfide. So now these moles cancel out and that equals when we work this out, that's gonna give me 12.7 grams of chromium(III) sulfide. So chromium(III) oxide says that it can make this much product. Let's see how much hydrogen sulfide says it can make. We have 12.80 grams of hydrogen sulfide. We have 1 mole of hydrogen sulfide on top and then grams of hydrogen sulfide on the bottom. When we add together the 2 hydrogens, each one is 1.008 grams, and the 1 sulfur, which is 32.07 grams, their combined mass is 34.086 grams. Grams cancel out. Now I'm gonna go from moles of given to moles of unknown, And remember, when we make this jump we look at the coefficients. So coming back up here, for every 3 moles of hydrogen sulfide we have 1 of chromium(III) sulfide. So it's 3 then 1. Here, these moles cancel out and then I already calculated the mass for chromium(III) sulfide so just plug it back in. So when we do that, that's gonna give us 17 grams of chromium(III) sulfide. So we have 2 totals here. How do we determine which one is the correct answer? Well, that's when we look at step 4. Compare the final amounts of the unknown to determine the theoretical yield. Remember, theoretical yield represents your 100% yield, how much you can make maximum. We're gonna say here that the smaller amount represents my limiting reagent and the larger amount represents my excess reagent. So here are our 2 totals. The smaller amount is 12.7 grams. Those 12.7 grams were produced by this chromium(III) oxide. So chromium(III) oxide represents my limiting reagent. The larger amount of 17 grams was produced by hydrogen sulfide. Hydrogen sulfide represents my excess reagent. Remember, it is a limiting reagent that determines a theoretical yield. It is the one that determines how much product you could theoretically make if this reaction was working perfectly. So the limiting reagent is chromium(III) oxide, and so its total amount of product it says it can make is our theoretical yield. So this would be the answer here. This is how much product we could potentially make if everything was working to a 100% efficiency. Right? So just remember, if they're giving you multiple grams of given, moles of given, any units of given, find out how much product each compound can make. The answer to the theoretical yield is always the smaller answer. That smaller answer is created by the limiting reagent. Okay? So in this question, 12.7 grams of our product would be our final answer.

Acrylonitrile (C­₃H₃N) is the starting material for many synthetic carpets and fabrics. It is produced by the following reaction: 

2 C₃H₆ (g) + 2 NH₃ (g) + 3 O₂ (g)  → 2 C₃H₃N (g) + 6 H₂O (g)

If 12.0 g C₃H₆, 10.0 g NH­₃, and 5.0 g O₂ react, what mass of acrylonitrile can be produced, assuming 100% yield?

• A. 31.2 g
• B. 15.1 g
• C. 5.50 g
• D. 12.4 g
• E. 15.6 g

The reaction between solid aluminum and iron (III) oxide can generate temperatures reaching 3000 ºC and is used in welding metals. 

2 Al + Fe₂O₃ → Al₂O_­3 + 2 Fe

If 150 g of Al are reacted with 432 g of Fe­₂O₃, what is the mass of the excess reactant remaining? 

• A. 2.21 g
• B. 8.08 g
• C. 11.2 g
• D. 4.04 g
• E. 7.75 g

Here’s what students ask on this topic:

What is a limiting reagent in a chemical reaction?

A limiting reagent, also known as a limiting reactant, is the reactant in a chemical reaction that is completely consumed first. This reactant determines the maximum amount of product that can be formed, known as the theoretical yield. Once the limiting reagent is used up, the reaction cannot proceed further, even if other reactants are still available. Identifying the limiting reagent is crucial for calculating the theoretical yield and understanding the efficiency of the reaction.

How do you determine the limiting reagent in a chemical reaction?

To determine the limiting reagent, follow these steps: 1) Write the balanced chemical equation. 2) Convert the given amounts of each reactant to moles. 3) Use stoichiometry to calculate the amount of product each reactant can produce. 4) The reactant that produces the least amount of product is the limiting reagent. This process involves comparing the mole ratios from the balanced equation to the actual amounts of reactants used.

What is the difference between a limiting reagent and an excess reagent?

The limiting reagent is the reactant that is completely consumed in a chemical reaction, determining the maximum amount of product that can be formed. In contrast, the excess reagent is the reactant that remains after the reaction has completed. The excess reagent is not fully used up because there is more of it than needed to react with the limiting reagent. Identifying both helps in calculating the theoretical yield and understanding the reaction's efficiency.

Why is it important to identify the limiting reagent in a chemical reaction?

Identifying the limiting reagent is important because it determines the maximum amount of product that can be formed in a chemical reaction, known as the theoretical yield. This information is crucial for predicting reaction outcomes, optimizing reactant usage, and minimizing waste. It also helps in scaling up reactions for industrial applications and ensuring that reactions are cost-effective and efficient.

Can a chemical reaction have more than one limiting reagent?

No, a chemical reaction cannot have more than one limiting reagent. By definition, the limiting reagent is the single reactant that is completely consumed first, limiting the amount of product formed. While multiple reactants may be present in limited quantities, only one will be the limiting reagent that determines the theoretical yield of the reaction.