# 12.3: Limiting Reactant and Percent Yield

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

• Define limiting reagent, theoretical yield, and percent yield.
• Calculate theoretical yield for a given chemical process.
• Be able to determine which reactant is the limiting reactant, calculate the amount of product formed, and determine the percent yield.
• Use reaction tables to describe mass and mole changes for a given chemical reaction.

## Lesson Vocabulary

• actual yield: The amount of product that is actually produced.
• theoretical yield: The maximum amount of product that can be generated from the given amounts of reactants.
• percent yield: Tells us what percentage of the possible amount of product (the theoretical yield) was actually obtained (the actual yield).
• excess reactant (excess reagent): When there is more reactant available than is required to react with the other available reactants. Some amount of reactant will be leftover at the end of the reaction.
• limiting reactant (limiting reagent): The reactant that is completely consumed in a reaction.

Write the balanced chemical equations for the following reactions:

1. Methane (CH4) reacts with oxygen in the air to produce water and carbon dioxide.
2. Solutions of barium chloride and sodium sulfate react to form a precipitate of barium sulfate and an aqueous solution of sodium chloride.
3. Carbon monoxide reacts with oxygen to produce carbon dioxide.

## Introduction

In the last lesson, we learned how to perform stoichiometry calculations, which relate masses and moles of reactants and products for a given chemical process. In this lesson, we are going to compare theoretical yield (the maximum amount that could be produced in a reaction) to actual yield (the amount that is actually produced). We will also investigate what happens when one reactant runs out before the other reactants are fully consumed. Finally, we will study how to express changes in masses and moles for a given chemical process using the reaction table method.

## Reaction Yield

### Actual vs. Theoretical Yield

The yield of a chemical reaction is the amount of certain product that is produced from given amounts of each reactant. The actual yield is the amount of product that is actually produced. This value is generally not exactly equal to the theoretical yield, which represents the maximum amount that could be generated from the given amounts of reactants. For example, say we performed the copper cycle as described in the introduction by starting with 1.00 grams of copper. Theoretically, the final reaction should give us back 1.00 grams of copper; this is our theoretical yield. However, we may find that only 0.86 grams of Cu is produced; this would be our actual yield. When we perform stoichiometric calculations, we are attempting to determine the theoretical yield based on the amounts of reactants available. Actual yields can only be determined by performing the experiment and measuring the final mass of product.

### Percent Yield

A common way to express the yield of a reaction is as a percentage. The percent yield of a reaction tells us what percentage of the possible amount of product (the theoretical yield) was actually obtained (the actual yield). Percent yield can be calculated using the following expression:

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

Example 12.6

You calculate that 1.00 grams of copper should be produced (theoretical yield) from a given chemical process. After you run the experiment, you find that 0.860 grams of copper is obtained. Calculate the percent yield for this process.

\begin{align*}\text{Percent Yield} &= \mathrm{\frac{0.860 \ g \ Cu}{1.00 \ g \ Cu}} \times 100\% \\ &=86\% \end{align*}

## The Reaction Table Method

A reaction table can be used to keep track of the masses and moles of each reaction component over the course of a chemical reaction. For the generic reaction shown below, we can set up the following Table below.

aA + bB → cC + dD
A B C D
Molar Mass
Initial Mass
Initial Moles
Change in Moles -ax -bx +cx +dx
Final Moles
Final Mass

Note: The change in the number of moles for each reactant and product must be consistent with the mole ratios in the balanced chemical equation. This is designated by the factor of "x" in the Change in Moles row of the reaction table. Also recall that as a chemical reaction takes place, the reactants are being used up (indicating a negative change in moles) and the products are being created (indicating a positive change in moles).

Steps to Solving Chemical Reaction Problems

1. Write the balanced reaction.
2. Draw a reaction table.
3. Fill in the known values.
4. Calculate the missing values.

This process is easiest to explain in the form of an example problem.

Example 12.7

Magnesium metal is heated in the presence of oxygen gas to produce magnesium oxide.

2Mg(s)+ O2(g) → 2MgO(s)

If 1.00 grams of Mg react completely with excess oxygen, how many grams of magnesium oxide will be produced? How many grams of oxygen will be used?

First, write the balanced reaction, and then draw a blank reaction table. The changes in moles of each component can be written in terms of a variable x, which we will solve for. The relative number of moles added to or subtracted from each amount is based on the coefficients from the balanced equation.

2Mg(s) + O2(g) → 2MgO(s)
Mg O2 MgO
Molar Mass
Initial Mass
Initial Moles
Change in Moles -2x -x +2x
Final Moles
Final Mass

Now, fill in the known values. We are told that the initial mass of magnesium is 1.00 grams. We are also told that our magnesium sample reacts completely, so the final mass of magnesium (and the final moles) will be 0. We do not know exactly how much oxygen gas is present, but it is an excess reactant (sometimes called an excess reagent), which means that there is more than enough oxygen to react with the other available reactants. We can simply write "excess" in both the initial and final masses/moles for this reactant. Unless told otherwise, assume that there is no initial product (the initial mass and moles of MgO would be 0). Additionally, we can calculate the molar masses for each reactant and product by looking at the periodic table:

Mg = 24.31 g/mol
O2 = 32.00 g/mol
MgO = 40.31 g/mol
Mg O2 MgO
Molar Mass 24.31 g/mol 32.00 g/mol 40.31 g/mol
Initial Mass 1.00 g excess 0 g
Initial Moles excess 0 mol
Change in Moles -2x -x +2x
Final Moles 0 mol excess
Final Mass 0 g excess

Now, fill in the remaining values by performing calculations. We can use molar masses to convert between grams and moles for any known amount.

\begin{align*}1.00 \text{ g Mg} (\mathrm{\frac{1 \ mol \ Mg}{24.31 \ g \ Mg}}) = 0.0411 \text{ mol Mg}\end{align*}

Once we know the difference between the initial and final moles for any of the reaction components, we can calculate the value of x. Over the course of the reaction, 0.0411 moles of Mg are used up, so

\begin{align*}-2x &= -0.0411 \ \text{mol} \\ x &= 0.0206 \ \text{mol}\end{align*}

Using this information, we can fill in more of the table:

Mg O2 MgO
Molar Mass 24.31 g/mol 32.00 g/mol 40.31 g/mol
Initial Mass 1.00 g excess 0 g
Initial Moles 0.0411 mol excess 0 mol
Change in Moles -0.0411 mol -0.0206 mol +0.0411 mol
Final Moles 0 mol excess
Final Mass 0 g excess

Using our value for x, we can now perform simple addition to determine the final moles of MgO. This can then be converted to grams using the molar mass.

Mg O2 MgO
Molar Mass 24.31 g/mol 32.00 g/mol 40.31 g/mol
Initial Mass 1.00 g excess 0 g
Initial Moles 0.0411 mol excess 0 mol
Change in Moles -0.0411 mol -0.0206 mol +0.0411 mol
Final Moles 0 mol excess 0.0411 mol
Final Mass 0 g excess 1.66 g

Now that the table is complete, we can answer the original questions. We can see directly from the table that 1.66 grams of MgO could be produced from this reaction. We do not yet know the mass of oxygen that is used up, but we do know that 0.0206 moles are consumed. Converting this to grams using the molar mass of O2 gives us the following:

\begin{align*}0.0206 \ \text{mol O}_2 \times (\mathrm{\frac{32.00 \ g \ O_2}{1 \ mol \ O_2}}) = 0.659 \ \text{g O}_2\end{align*}

1.00 grams of Mg reacts completely with 0.659 grams of O2 to produce 1.66 grams of MgO.

## Limiting Reactant

So far, we have either assumed that all reactants will be completely consumed in a chemical reaction, or we have been told that one reactant is present in excess. In the previous example, 1 gram of magnesium reacted in the presence of excess oxygen. However, we may sometimes be presented with initial amounts of multiple reactants without being told which one will run out first. In such a scenario, the limiting reactant (sometimes called a limiting reagent) will be the reactant that is completely consumed. After the limiting reactant runs out, there may still be some of the excess reactants left over, but the reaction can no longer proceed, because one of the ingredients is missing. To determine which reactant is limiting in a chemical reaction, we need to look at how many moles of each are present. These values must then be compared to the coefficients in the balanced equation, which tell us the ratios in which various reactants combine.

Before working with chemical reactions, it may help to explain the concept of limiting reactants in a more familiar context. For example, let's say that you want to make as many cheese sandwiches as possible with the bread and cheese that is available.

Example 12.8

You have 16 slices of bread and 10 slices of cheese. If each sandwich requires two slices of bread and one slice of cheese, how many sandwiches can you make, and what ingredients will be left over? For this "reaction," which reactant is limiting, and which one is present in excess?

For this example, we could simply start subtracting bread and cheese as each sandwich is made. After making 8 sandwiches, we would find that we have run out of bread, but there are two slices of cheese left over. Thus, bread is the limiting reactant, and cheese is present in excess. Notice that we actually have more slices of bread than of cheese, but because it gets used up twice as fast, bread runs out first (it is limiting).

An alternative way to look at this problem would be to write this "reaction" out as a chemical equation.

2 Bread + Cheese → Sandwich

We cannot directly compare the amounts of bread and cheese, because they are not used in a 1:1 ratio. However, if we divide each amount by the coefficient from the balanced equation, we get the following:

\begin{align*}\mathrm{\frac{16 \ slices \ bread}{2}} &=8 \\ \mathrm{\frac{10 \ slices \ cheese}{1}} &=10\end{align*}

These values can be directly compared. After dividing each amount by its coefficient from the balanced equation, the smallest number corresponds to the ingredient that will run out first. In this case, the limiting reactant is bread, because 8 < 10. To determine how much of the excess reactant is left over, we need to determine how much will be used up by the limiting reactant. In order to use up all of the limiting reactant (16 slices of bread), we would need:

\begin{align*}16 \ \text{slices bread} \times (\mathrm{\frac{1 \ slice \ cheese}{2 \ slices \ bread}}) = 8 \ \text{slices cheese}\end{align*}

After using 8 slices of cheese, all the bread will be used up, and the "reaction" stops. Subtracting this amount from our original 10 slices of cheese, we can see that there will be two slices left over.

## Limiting Reactants and Reaction Tables

Filling out a reaction table will help to double-check that you have found the correct limiting reactant, and it will give you information about theoretical yields and the amounts that will be left over for any excess reactants.

Example 12.9

3.40 g of hydrogen gas and 7.16 g of nitrogen gas react to form gaseous ammonia. Assume that the reaction runs until one of the reactants is fully consumed.

First, write the balanced equation for this reaction and set up a reaction table. Fill in the amounts given in the problem, and calculate the molar masses of each reaction component. Recall that the "change in moles" line is based on the coefficients from the balanced equation:

N2(g) + 3 H2(g) → 2NH3(g)
N2 3H2 2NH3
Molar Mass 28.02 g/mol 2.02 g/mol 17.04 g/mol
Initial Mass 7.16 g 3.40 g 0 g
Initial Moles 0 mol
Change in Moles -x -3x +2x
Final Moles
Final Mass

Next, use molar masses to convert the known masses to moles:

\begin{align*}7.16 \ \text{g N}_2 (\mathrm{\frac{1 \ mol \ N_2}{28.02 \ g \ N_2}}) &= 0.256 \ \text{mol N}_2 \\ 3.40 \ \text{g H}_2 (\mathrm{\frac{1 \ mol \ H_2}{2.02 \ g \ H_2}}) &= 1.68 \ \text{mol H}_2\end{align*}

Now that we have the moles of each reactant, we can divide by the coefficients from the balanced equation to determine which will run out first.

\begin{align*}\mathrm{\frac{0.256 \ mol \ N_2}{1}} &=0.256 \\ \mathrm{\frac{1.68 \ mol \ H_2}{3}} &=0.560\end{align*}

Because the number for N2 is lower, N2 is the limiting reactant. Since this reaction is run until a reactant runs out, we now know that the final mass of N2 will be 0 grams. Putting this new information into the table, we get the following:

N2 3H2 2NH3
Molar Mass 28.02 g/mol 2.02 g/mol 17.04 g/mol
Initial Mass 7.16 g 3.40 g 0 g
Initial Moles 0.256 mol 1.68 mol 0 mol
Change in Moles -x -3x +2x
Final Moles 0 mol
Final Mass 0 g

Looking at the difference between the initial and final moles for N2, it is clear that x must have a value of 0.256 moles. Using this value, we can replace the "Change in Moles" line with specific values, and the "Final Moles" line can be filled in by simple addition or subtraction.

N2 3H2 2NH3
Molar Mass 28.02 g/mol 2.02 g/mol 17.04 g/mol
Initial Mass 7.16 g 3.40 g 0 g
Initial Moles 0.256 mol 1.68 mol 0 mol
Change in Moles -0.256 mol -0.768 mol +0.512 mol
Final Moles 0 mol 0.91 mol 0.512 mol
Final Mass 0 g

Finally, we can convert moles to grams using the molar masses of each reaction component:

\begin{align*}0.91 \ \text{mol H}_2 \times (\mathrm{\frac{2.02 \ g \ H_2}{1 \ mol \ H_2}}) &=1.84 \ \text{g H}_2 \\ 0.512 \ \text{mol NH}_3 \times (\mathrm{\frac{17.04 \ g \ NH_3}{1 \ mol \ NH_3}}) &=8.72 \ \text{g NH}_3\end{align*}

N2 3H2 2NH3
Molar Mass 28.02 g/mol 2.02 g/mol 17.04 g/mol
Initial Mass 7.16 g 3.40 g 0 g
Initial Moles 0.256 mol 1.68 mol 0 mol
Change in Moles -0.256 mol -0.768 mol +0.512 mol
Final Moles 0 mol 0.91 mol 0.512 mol
Final Mass 0 g 1.84 g 8.72 g

Note that if we calculate the total mass of all components, it is the same before and after the reaction (10.56 g total). This is consistent with the law of conservation of mass.

Now, we can answer specific questions about this reaction. The limiting reactant is N2, because it is completely used up before any of the other reactants. We would know that we chose the wrong limiting reactant if any of the other final amounts turned out to be negative numbers. The theoretical yield of ammonia is 8.72 g, and there is 1.84 g of H2 left over after the reaction is complete.

If the yield is 87.5%, we can perform the following calculation:

\begin{align*} \mathrm{Percent \ Yield} &= \mathrm{\frac{Actual \ Yield}{Theoretical \ Yield} \times 100 \%} \\ \mathrm{Actual \ Yield} &= \mathrm{(Theoretical \ Yield)(\frac{Percent \ Yield}{100 \%})} \\ \mathrm{Actual \ Yield} &= \mathrm{(8.62 \ g)(\frac{87.5 \%}{100 \%})} \\ \mathrm{Actual \ Yield} &= \mathrm{7.54 \ g}\end{align*}

7.54 grams of ammonia were actually produced in this reaction.

## Lesson Summary

• Chemical reactions can be studied in terms of the amounts of reactants that are present and the amount of products that are formed.
• The balanced chemical equation describes the relative amounts of reactants and products that are consumed and generated by a given chemical process.
• Reactants combine in fixed mole ratios.
• The amount of each reactant needed to generate a certain amount of product can be determined with stoichiometric calculations.
• In cases where one reactant is used up before the others, we can describe the reaction in terms of a limiting reactant and one or more excess reactants.
• The actual and theoretical yields for a given reaction are often related by calculating a percent yield.
• The reaction table method provides a way to organize the information needed to perform stoichiometry calculations and to answer quantitative questions about various chemical reactions.

## Lesson Review Questions

1. Urea (NH2)2CO is prepared by reacting ammonia with carbon dioxide according to the following equation: 2NH3(g) + CO2(g) → (NH2)2CO(aq) + H2O(l). In one experiment, 637.2 g of NH3 is allowed to react with 1142 g of CO2.
1. Which of the two reactants is the limiting reagent?
2. Calculate the mass of(NH2)2CO that could theoretically be formed by this reaction.
2. Hydrogen gas reacts explosively in the presence of oxygen to produce water.
1. Write the balanced chemical equation for this process.
2. If 1 mol of hydrogen gas and 1 mol of oxygen gas are placed in a container and ignited, how many moles of water will be produced?
3. Which will be the limiting reactant?
4. Which will be the excess reactant, and how much of the excess reactant will be left over?
3. If 1.00 grams of hydrogen gas and 1.00 grams of oxygen gas are placed in a container and ignited:
1. Which will be the limiting reactant?
2. Which will be the excess reactant?
3. What is the theoretical yield for water?
4. If 0.96 g of water are actually produced, what is the percent yield for this process?
4. Gaseous nitrogen and hydrogen react to form ammonia via the Haber process, which is represented by the following chemical equation: N2(g) + 3H2(g) → 2NH3(g)
1. If you react 25.0 grams of hydrogen gas in the presence of excess nitrogen gas, how many grams of ammonia should be produced?
2. If you produced 140. grams of ammonia, what is the percent yield?
5. Look back to Example 12.8. If you changed the recipe to 2 slices of cheese per sandwich, what would be the limiting reactant and excess reactant given the same reactant amounts as in the example?
6. Magnesium reacts with nitrogen in the air to form magnesium nitride.
1. Write and balance the chemical equation.
2. What is the limiting reactant if 4.00 g of Mg are mixed with 8.00 g of nitrogen?
3. What is the theoretical yield of magnesium nitride in grams?

## Points to Consider

1. What are some factors that control whether or not the actual yield of a chemical reaction is 100%? What would it mean for a reaction yield to be less than 100%? More than 100%?

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