<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# 12.3: Limiting Reactant and Percent Yield

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

• Analyze a chemical reaction in order to determine which reactant is the limiting reactant and which is the excess reactant.
• Calculate the amount of excess reactant remaining after a reaction is complete.
• Calculate the theoretical yield of a reaction when the available amounts of each reactant are known.
• Calculate the percent yield of a reaction based on the theoretical and actual yields.

## Lesson Vocabulary

• actual yield
• excess reactant (reagent)
• limiting reactant (reagent)
• percent yield
• theoretical yield

### Recalling Prior Knowledge

• What is the relationship between the amounts of reactants in a chemical reaction that follows ideal stoichiometry?
• How is the mass of a product determined when the mass of one reactant is known?

We have seen that the reactants in a chemical reaction combine with one another in a specific molar ratio that is described by the balanced equation. What if a reaction is performed in which the reactants are not present in that exact ratio? In this lesson, you will learn about this type of reaction and how to calculate the amounts of products that are expected to be formed.

## Identifying the Limiting Reactant

### Everyday Limiting Reactant

Cooking is a great example of everyday chemistry. In order to correctly follow a recipe, a cook needs to make sure that he has plenty of all the necessary ingredients in order to make his dish. Let us suppose that you are planning on making some pancakes for a large group of people (Figure below).

When making pancakes, the number that you can make is limited by the amount of each ingredient that you have.

The recipe on the box indicates that the following ingredients are needed for each batch of pancakes:

• 1 cup of pancake mix
• 34\begin{align*}\frac{3}{4}\end{align*} cup milk
• 1 egg
• 1 tablespoon vegetable oil

Now you check the pantry and the refrigerator and see that you have the following ingredients available:

• 2 boxes of pancake mix (8 cups)
• Half gallon of milk (4 cups)
• 2 eggs
• Full bottle of vegetable oil (about 3 cups)

The question that you must ask is: How many batches of pancakes can I make? The answer is two. Even though you have enough pancake mix, milk, and oil to make many more batches of pancakes, you are limited by the fact that you only have two eggs. As soon as you have made two batches of pancakes, you will be out of eggs and your “reaction” will be complete.

### Limiting Reactants in Chemical Reactions

For a chemist, the balanced chemical equation is the recipe that must be followed. As you have seen earlier, the Haber process is a reaction in which nitrogen gas is combined with hydrogen gas to form ammonia. The balanced equation is shown below.

N2(g) + 3H2(g) → 2NH3(g)

We know that the coefficients of the balanced equation tell us the mole ratio that is required for this reaction to occur. One mole of N2 will react with three moles of H2 to form two moles of NH3.

Now let us suppose that a chemist were to react three moles of N2 with six moles of H2.

What happened in this reaction? The chemist started with 3 moles of N2. You may think of this as being enough to make 3 batches of the “recipe” (as shown in the balanced equation), since the coefficient for N2 is 1. However, the 6 moles of H2 that the chemist started with is only enough for 2 batches, since the coefficient for H2 is 3, and 3 × 2 = 6. After this reaction is complete, all of the hydrogen gas will be gone, but there will be 1 mole of nitrogen gas left over. At most, 4 moles of NH3 (two "batches") can be generated with these amounts of reactants, because after that point, one of the "ingredients" has been completely used up. The overall reaction that occurred is the following:

2 mol N2 + 6 mol H2 → 4 mol NH3

All the amounts are doubled from the original balanced equation.

The limiting reactant (or limiting reagent) is the reactant that determines the amount of product that can be formed in a chemical reaction. The reaction proceeds until the limiting reactant is completely used up. In our example above, H2 is the limiting reactant. The excess reactant (or excess reagent) is any reactant that cannot be completely consumed by the full reaction of the limiting reactant. In other words, there is always excess reactant left over after the reaction is complete. In the above example, N2 is the excess reactant.

Get help with limiting reactant problems at http://www.chem.tamu.edu/class/majors/tutorialnotefiles/limiting.htm.

Watch an animation of a reaction involving a limiting reactant at http://www.dlt.ncssm.edu/core/Chapter6-Stoichiometry/Chapter6-Animations/LimitingReactant.html.

### Solving Limiting Reactant Problems

In the real world, amounts of reactants and products are typically measured by mass or by volume. It is first necessary to convert the given quantities of each reactant to moles in order to identify the limiting reactant.

Sample Problem 12.9: Determining the Limiting Reactant

Silver metal reacts with sulfur to form silver sulfide according to the following balanced equation:

2Ag(s) + S(s) → Ag2S(s)

What is the limiting reactant when 50.0 g Ag is reacted with 10.0 g S?

Step 1: List the known quantities and plan the problem.

Known

• given: 50.0 g Ag
• given: 10.0 g S

Unknown

• limiting reactant

Use the atomic masses of Ag and S to determine the number of moles of each present. Then, use the balanced equation to calculate the number of moles of sulfur that would be needed to react with the number of moles of silver present. Compare this result to the actual number of moles of sulfur present.

Step 2: Solve.

First, calculate the number of moles of Ag and S present:

50.0 g Ag×1 mol Ag107.87 g Ag=0.464 mol Ag\begin{align*}50.0 \ \text{g Ag} \times \frac{1 \ \text{mol Ag}}{107.87 \ \text{g Ag}} = 0.464 \ \text{mol Ag}\end{align*}
10.0 g S×1 mol S32.07 g S=0.312 mol S\begin{align*}10.0 \ \text{g S} \times \frac{1 \ \text{mol S}}{32.07 \ \text{g S}} = 0.312 \ \text{mol S}\end{align*}

Second, find the moles of S that would be required to react with all of the given Ag:

0.464 mol Ag×1 mol S2 mol Ag=0.232 mol S (required)\begin{align*}0.464 \ \text{mol Ag} \times \frac{1 \ \text{mol S}}{2 \ \text{mol Ag}} = 0.232 \ \text{mol S (required)}\end{align*}

The amount of S actually present is 0.312 moles. The amount of S that is required to fully react with all of the Ag is 0.232 moles. Since there is more sulfur present than what is required to react, the sulfur is the excess reactant. Therefore, silver is the limiting reactant.

The balanced equation indicates that the necessary mole ratio of Ag to S is 2:1. Since there were not twice as many moles of Ag present in the original amounts, that makes silver the limiting reactant.

There is a very important point to consider about the preceding problem. Even though the mass of silver present in the reaction (50.0 g) was greater than the mass of sulfur (10.0 g), silver was the limiting reactant. This is because chemists must always convert to molar quantities and consider the mole ratio from the balanced chemical equation.

Practice Problems
1. The equation for the combustion of propane (C3H8) is: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g). If 1.30 mol C3H8 is reacted with 6.00 mol O2, identify the limiting reactant.
2. When zinc metal reacts with hydrochloric acid, zinc chloride and hydrogen gas are produced: Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g). Identify the limiting reactant when 8.00 g Zn reacts with 8.00 g HCl.

There are two other things that we would like to be able to determine in a limiting reactant problem. One is the quantity of the excess reactant that will be left over after the reaction is complete. Another is the quantity of product that will be formed in the reaction. We will go back to Sample Problem 12.9 to answer these questions.

Sample Problem 12.10A: Determining the Amount of Excess Reactant Left Over

What is the mass of excess reactant remaining when 50.0 g Ag reacts with 10.0 g S?

2Ag(s) + S(s) → Ag2S(s)

Step 1: List the known quantities and plan the problem.

Known

• Excess reactant = 0.312 mol S (from Sample Problem 12.9)
• Amount of excess reactant needed = 0.232 mol S (from Sample Problem 12.9)

Unknown

• Mass of excess reactant remaining after the reaction = ? g

Subtract the amount (in moles) of the excess reactant that will react from the amount that is originally present. Convert moles to grams.

Step 2: Solve.

0.312 mol S0.232 mol S=0.080 mol S (remaining after reaction)\begin{align*}0.312 \ \text{mol S} - 0.232 \ \text{mol S} = 0.080 \ \text{mol S (remaining after reaction)}\end{align*}
0.080 mol S×32.07 g S1 mol S=2.57 g S\begin{align*}0.080 \ \text{mol S} \times \frac{32.07 \ \text{g S}}{1 \ \text{mol S}} = 2.57 \ \text{g S}\end{align*}

There are 2.57 g of sulfur remaining when the reaction is complete.

There were 10.0 g of sulfur present before the reaction began. If 2.57 g of sulfur remains after the reaction, then 7.43 g S reacted.

7.43 g S×1 mol S32.07 g S=0.232 mol S\begin{align*}7.43 \ \text{g S} \times \frac{1 \ \text{mol S}}{32.07 \ \text{g S}} = 0.232 \ \text{mol S}\end{align*}

This is the amount of sulfur that reacted. The problem is internally consistent.

Sample Problem 12.10B: Determining the Quantity of Product Formed in a Reaction

What mass of Ag2S will be produced when 50.0 g Ag reacts with 10.0 g S?

2Ag(s) + S(s) → Ag2S(s)

Step 1: List the known quantities and plan the problem.

Known

• Limiting reactant = 0.464 mol Ag (from Sample Problem 12.9)
• Molar mass of Ag2S = 247.81 g/mol

Unknown

• Mass of silver sulfide produced = ? g

The limiting reactant is the reactant that will determine the amount of product that is produced. Use stoichiometry to calculate the number of moles of Ag2S produced and then convert that amount to grams.

mol Ag → mol Ag2S → g Ag2S

Step 2: Solve.

0.464 mol Ag×1 mol Ag2S2 mol Ag×247.81 g Ag2S1 mol Ag2S=57.5 g Ag2S\begin{align*}0.464 \ \text{mol Ag} \times \dfrac{1 \ \text{mol Ag}_2\text{S}}{2 \ \text{mol Ag}} \times \dfrac{247.81 \ \text{g Ag}_2\text{S}}{1 \ \text{mol Ag}_2\text{S}} = 57.5 \ \text{g Ag}_2\text{S}\end{align*}

There is 57.5 g Ag2S formed in the reaction.

Silver is the limiting reactant, so all 50.0 grams of silver are used up in the reaction. Therefore, the mass of Ag2S produced must be larger than 50.0 g. However, the mass of Ag2S produced must be less than 60.0 g, because the sulfur was in excess, so some of it did not react.

Practice Problems
1. Sodium chloride can be produced by reacting solid sodium with chlorine gas: In a certain reaction, 3.40 moles of Na are reacted with 4.20 moles of Cl2.
1. Identify the limiting reactant.
2. Calculate the moles of excess reactant remaining after the reaction.
3. Calculate the moles of NaCl produced.
2. Using the Haber process for the production of ammonia: How many grams of ammonia will be produced when 35.0 g N2 is reacted with 12.0 g H2?

View another example of solving a limiting reactant problem at http://www.khanacademy.org/science/physics/thermodynamics/v/limiting-reactant-example-problem-1.

## Percent Yield

Chemical reactions in the real world don’t always go exactly as planned on paper. In the course of an experiment, many things will contribute to the formation of less product than would be predicted. Besides spills and other experimental errors, there are usually losses due to incomplete reactions, undesirable side reactions, and numerous other factors. Chemists need a measurement that indicates how successful a reaction has been. This measurement is called the percent yield.

To compute the percent yield, it is first necessary to determine how much of the product should be formed based on stoichiometry. This is called the theoretical yield, the maximum amount of product that could be formed from the given amounts of reactants. The actual yield is the amount of product that is actually formed when the reaction is carried out in the laboratory. The percent yield is the ratio of the actual yield to the theoretical yield, expressed as a percentage.

Percent Yield=Actual YieldTheoretical Yield×100%\begin{align*}\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%\end{align*}

Percent yield is very important in the manufacture of products. Much time and money is spent improving the percent yield for chemical production. When complex chemicals are synthesized by many different reactions, one step with a low percent yield can quickly cause a large waste of reactants and money.

Percent yields are generally less than 100% because of the reasons indicated earlier. However, percent yields that appear to be greater than 100% are possible if the measured product of the reaction contains impurities that add to the mass of the pure product. When a chemist synthesizes a desired chemical, he or she must always be careful to purify the products of the reaction.

Sample Problem 12.11: Calculating the Theoretical Yield and the Percent Yield

Potassium chlorate decomposes when heated in the presence of a catalyst according to the reaction below:

2KClO3(s) → 2KCl(s) + 3O2(g)

In a certain experiment, 40.0 g KClO3 is heated until it completely decomposes. What is the theoretical yield of oxygen gas? The experiment is performed, and the mass of the collected oxygen gas is found to be 14.9 g. What is the percent yield for the reaction?

Part A: First, we will calculate the theoretical yield based on the stoichiometry.

Step 1: List the known quantities and plan the problem.

Known

• given: mass of KClO3 = 40.0 g
• molar mass of KClO3 = 122.55 g/mol
• molar mass of O2 = 32.00 g/mol

Unknown

• theoretical yield of O2 = ? g

Apply stoichiometry to convert from the mass of a reactant to the mass of a product:

g KClO3 → mol KClO3 → mol O2 → g O2

Step 2: Solve.

40.0 g KClO3×1 mol KClO3122.55 g KClO3×3 mol O22 mol KClO3×32.00 g O21 mol O2=15.7 g O2\begin{align*}40.0 \ \text{g KClO}_3 \times \frac{1 \ \text{mol KClO}_3}{122.55 \ \text{g KClO}_3} \times \frac{3 \ \text{mol O}_2}{2 \ \text{mol KClO}_3} \times \frac{32.00 \ \text{g O}_2}{1 \ \text{mol O}_2} = 15.7 \ \text{g O}_2\end{align*}

The theoretical yield of O2 is 15.7 g.

The mass of oxygen gas must be less than the 40.0 g of potassium chlorate that was decomposed.

Part B: Now we use the actual yield and the theoretical yield to calculate the percent yield.

Step 1: List the known quantities and plan the problem.

Known

• Actual yield = 14.9 g
• Theoretical yield = 15.7 g (from Part A)

Unknown

• Percent yield = ? %
Percent Yield=Actual YieldTheoretical Yield×100%\begin{align*}\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%\end{align*}

Use the percent yield equation above.

Step 2: Solve.

Percent Yield=14.9 g15.7 g×100%=94.9%\begin{align*}\text{Percent Yield} = \frac{14.9 \ \text{g}}{15.7 \ \text{g}} \times 100\% = 94.9\%\end{align*}

Since the actual yield is slightly less than the theoretical yield, the percent yield is just under 100%.

Practice Problems
1. If 23.5 g of lithium nitride is reacted with an excess of water, 40.2 g of lithium hydroxide is produced according to the following reaction: What is the percent yield of the reaction?
2. Under a certain set of conditions, the Haber process is known to proceed such that the percent yield for the reaction is 86.0%. If the actual yield is 27.8 g NH3, what is the theoretical yield for the reaction?

Practice more percent yield problems at http://www.uen.org/utahlink/tours/tourElement.cgi?element_id=42288&tour_id=17891&category_id=33176.

You can find more practice problems at http://science.widener.edu/svb/tutorial/percentyieldcsn7.html.

## Lesson Summary

• The limiting reactant in a chemical reaction is the reactant that determines the amount of product that can be formed. More of the excess reactant is present than is needed to completely react with the limiting reactant, so some of the excess reactant remains after the reaction is complete.
• The theoretical yield is the maximum amount of product that could be formed based on stoichiometry calculations.
• The percent yield is a measure of the efficiency of the reaction and is calculated by dividing the actual yield by the theoretical yield and converting to a percentage.

## Lesson Review Questions

### Reviewing Concepts

1. What is wrong with this statement? “The limiting reactant of a reaction is whichever reactant is present in a smaller amount.”
2. Which reactant, limiting or excess, determines the theoretical yield of products?
3. Why are actual yields generally less than theoretical yields?
4. What is likely to have occurred if a calculated percent yield is larger than 100%?

### Problems

1. In a certain reaction, 0.740 mol of lithium is combined with 0.180 mol of nitrogen gas. The reaction forms lithium nitride according to the following equation:
1. Identify the limiting reactant.
2. Calculate the number of moles of excess reactant remaining after the reaction is complete.
3. Calculate the number of moles of product formed.
2. The reaction of fluorine gas with ammonia produces dinitrogen tetrafluoride and hydrogen fluoride. In a certain reaction, 45.8 g F2 is reacted with 20.6 g NH3.
1. Determine the limiting reactant.
2. How many grams of the excess reactant remains after the reaction is complete?
3. What is the theoretical yield of each of the two products?
4. The reaction is performed and 16.2 g of N2F4 is produced. Calculate the percent yield for the reaction.
3. 26.3 g of potassium metal reacts with 4.89 L of oxygen gas at STP, forming potassium oxide.
1. Determine the limiting reactant.
2. How much of the excess reactant remains after the reaction is complete? If the excess reactant is oxygen, give its volume.
3. What mass of potassium oxide could theoretically be produced?
4. The percentage yield for the reaction of PCl3 with Cl2 to form PCl5 is 80.8%. What mass of PCl5 would be produced by the reaction of 37.8 g of PCl3 with excess Cl2?
5. If 26.7 g of nickel metal is reacted with 178 g of silver nitrate, what mass of silver metal will be produced? What is the limiting reactant?
6. The Ostwald Process is a series of three reactions that produces nitric acid from ammonia. 1.00 kg of ammonia is subjected to the reactions, and unlimited amounts of oxygen and water are available. If each reaction proceeds at a 92.5% yield, what mass of HNO3 is produced? Assume that any NO produced by the third reaction is lost and not reused as a reactant in the second reaction.

## Points to Consider

Now that you have an understanding of the quantitative relationships that exist between chemical substances in chemical reactions, we will turn our attention to the states of matter and their relationships to one another.

• What are the three states of matter?
• How can changes in temperature and pressure affect the state of a sample of matter?

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes

Show Hide Details
Description
Tags:
Subjects: