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# Determining the Limiting Reactant

## Use molar ratios to determine which reactant will run out first in a reaction

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Determining the Limiting Reactant

Credit: Courtesy of Renee Comet and the National Cancer Institute
Source: http://commons.wikimedia.org/wiki/File:Brownie_%281%29.jpg

#### Who’s coming for dinner?

You have ten people that show up for a dinner party. One of the guest brings twenty brownies for dessert. The decision about serving dessert is easy: two brownies are placed on every plate. If someone wants more brownies, they will have to wait until they go to the store. There are only enough brownies for everyone to have two.

### Determining the Limiting Reactant

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: 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)\begin{align*}2\text{Ag}(s)+\text{S}(s) \rightarrow \text{Ag}_2\text{S}(s)\end{align*}

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 Ag10.0 g S×1 mol S32.07 g S=0.464 mol Ag=0.312 mol S\begin{align*}50.0 \text{ g Ag} \times \frac{1 \text{ mol Ag}}{107.87 \text{ g Ag}} &= 0.464 \text{ mol Ag}\\ 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} \ (\text{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.

There is one other thing that we would like to be able to determine in a limiting reactant problem - the quantity of the excess reactant that will be left over after the reaction is complete. We will go back to the sample problem above to answer this question.

#### Sample Problem: 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)\begin{align*}2\text{Ag}(s)+\text{S}(s) \rightarrow \text{Ag}_2\text{S}(s)\end{align*}

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 S0.080 mol S×32.07 g S1 mol S=0.080 mol S (remaining after reaction)=2.57 g S\begin{align*}0.312 \text{ mol S}-0.232 \text{ mol S} &= 0.080 \text{ mol S (remaining after reaction)}\\ 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.

### Summary

• Determining the limiting reactant requires that all mass quantities first be converted to moles to evaluate the equation.

### Review

1. Why do all mass values need to be converted to moles before determining the limiting reactant?
2. If we used 0.7 moles Ag, would it still be the limiting reactant?
3. If we ran the reaction using the original amounts of Ag and S and had 5.22 grams S left over, what might we assume about the reaction?

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