Limiting Reagent - Online Tutor, Practice Problems & Exam Prep

In stoichiometry, the limiting reagent is the reactant that is completely consumed, determining the maximum theoretical yield of a product. The excess reagent remains after the reaction. To identify these, calculate the amount of product each reactant can produce by converting grams to moles and using coefficients from the balanced equation. This process allows for the determination of the limiting reagent and the theoretical yield, essential for understanding chemical reactions and their efficiencies.

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. 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. We start talking about limiting reactant or reagent when more than one reactant in our chemical equation is given a starting amount. 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. Different from the limiting reagent is the excess reagent. 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. 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 out 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 of given for it. Convert those grams of given to moles of 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 decide 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 states that chromium(III) oxide reacts with hydrogen sulfide, which is H₂S gas, to form chromium(III) sulfide and water. We have chromium(III) oxide reacting with 3 moles of hydrogen sulfides to produce chromium(III) sulfide and 3 moles of water. 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 are providing us with more than one gram of given. So we have 2 grams of given, and they are asking us to find grams of an unknown. Because they are giving us multiple grams of given, I am going to have to do stoichiometry for both of those amounts. So I will go through the whole process of the stoichiometric chart with 14.20 grams and then again with 12.80 grams. Once we get our answers for the amount of chromium(III) sulfide produced, we will discuss those two answers and which one is the final solution.

Let's set this up. We have 14.20 grams of chromium(III) oxide. Let's start with converting 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 it weighs. So, I have to look at the periodic table. 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 is 16.00 grams. This comes out to 103.992 grams, and then 48.00 grams for the 3 oxygens. So together, that is 151.992 grams as the mass for chromium(III) oxide. Now, grams cancel out. We're going to convert these moles of given into moles of unknown, which is chromium(III) sulfide. Remember, at this point, we utilize the coefficients from the balanced equation. So it's for every 1 mole of chromium(III) oxide, there is 1 mole of chromium(III) sulfide. 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. Now, let's calculate the mass for chromium(III) sulfide. So here we're going to have 3 sulfurs. Sulfur on the periodic table is approximately 32.070 grams. So multiplying that by 3 gives us 96.210 grams. Now, add the weight of 2 chromiums (103.992 grams), so that total is 200.202 grams of chromium(III) sulfide. When we work this out, that's going to give us 12.7 grams of chromium(III) sulfide. Now let's see how much hydrogen sulfide can make. So we have 12.80 grams of hydrogen sulfide, with 1 mole of hydrogen sulfide on top, and its grams on the bottom. With the 2 hydrogens each weighing 1.008 grams, and the 1 sulfur, which is 32.070 grams, their combined mass is 34.086 grams. Again, grams cancel out. Now, according to the coefficients, for every 3 moles of hydrogen sulfide, there is 1 mole of chromium(III) sulfide. So when we calculate that, it gives us 17 grams of chromium(III) sulfide. We have 2 totals here. How do we determine which one is the correct answer? Well, we compare the final amounts of the unknown to determine the theoretical yield. The smaller amount represents the limiting reagent, and the larger amount represents the excess reagent. Thus, the 12.7 grams of chromium(III) sulfide were produced by chromium(III) oxide, making it the limiting reagent. Hydrogen sulfide, which produced a larger amount of 17 grams, is the excess reagent. The limiting reagent determines the theoretical yield, which is the one that determines how much product could theoretically be produced if the reaction was working perfectly. So the answer here is that 12.7 grams of chromium(III) sulfide is the maximum amount that can be produced when everything is working to 100% efficiency, representing our theoretical yield. Therefore, in this question, 12.7 grams of our product would be the 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) chloride 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 chemistry?

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 from the reaction. 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, which is the maximum amount of product that can be produced based on the initial amounts of reactants.

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 for the reaction. 2) Convert the given amounts of reactants (usually in grams) to moles. 3) Use the stoichiometric coefficients from the balanced equation 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 reactant limits the reaction and determines the theoretical yield.

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 consumed because there is more of it than needed to react with the limiting reagent. Identifying both reagents is essential for understanding the reaction's efficiency and calculating the theoretical yield.

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 various applications, including industrial processes, laboratory experiments, and chemical manufacturing. Knowing the limiting reagent helps in optimizing reactant usage, minimizing waste, and improving the efficiency and cost-effectiveness of chemical processes.

How do you calculate the theoretical yield in a chemical reaction?

To calculate the theoretical yield, follow these steps: 1) Identify the limiting reagent by converting the given amounts of reactants to moles and using stoichiometric coefficients from the balanced equation. 2) Use the moles of the limiting reagent to calculate the moles of the desired product using the stoichiometric ratio. 3) Convert the moles of the product to grams (or another unit) using its molar mass. The result is the theoretical yield, which represents the maximum amount of product that can be formed from the given reactants.