Limiting Reactants: Unlocking the Key to Reaction Efficiency
- Consider you are baking cookies, and the recipe requires 2 cups of flour and 1 cup of sugar to make a batch.
- However, you only have 4 cups of flour but 3 cups of sugar. How many batches can you make?
- The answer depends on the ingredient that runs out first—in this case, the flour.
This idea of a "limiting ingredient" applies to chemistry as well, where reactions depend on the availability of reactants.
The Limiting Reactant and Excess Reactant
Limiting reagent
In a chemical reaction, the limiting reactant is the substance that is completely consumed first, stopping the reaction from proceeding further.
It determines the theoretical yield, or the maximum amount of product that can be formed.
Excess reactant
The excess reactant is the substance that remains after the reaction is complete because there’s more of it than needed.
Why Does This Matter?
- Understanding limiting reactants is essential for optimizing chemical reactions, whether in industrial processes, laboratory experiments, or environmental applications.
- It ensures efficient use of resources and minimizes waste.
Identifying the Limiting Reactant
To identify the limiting reactant, follow these steps:
- Write a Balanced Chemical Equation: Ensure the chemical equation is balanced to reflect the correct stoichiometric ratios of reactants and products.
- Calculate the Moles of Each Reactant: Use the given mass, volume, or concentration to determine the number of moles for each reactant.
- Compare the Mole Ratios:
- Use the balanced equation to determine how many moles of each reactant are required.
- Compare these ratios to the actual amounts available.
- Determine the Limiting Reactant: The reactant that provides fewer moles relative to the stoichiometric requirement is the limiting reactant.
- Always start with a balanced chemical equation.
- Without it, mole ratio comparisons will be inaccurate.
Let’s determine the limiting reactant in the combustion of butane, $ \text{C}_4\text{H}_{10} $, using the reaction:
$$
2 \text{C}_4\text{H}_{10} (g) + 13\text{O}_2(g) \rightarrow 8\text{CO}_2(g) + 10\text{H}_2\text{O}(l)
$$
In a reaction mixture, you have 0.20 mol of $ \text{C}_4\text{H}_{10} $ and 2.6 mol of $ \text{O}_2 $. Which reactant is limiting?
Solution
Step 1: Write the Mole Ratios
From the balanced equation:
- 2 moles of $ \text{C}_4\text{H}_{10} $ react with 13 moles of $ \text{O}_2 $.
- The mole ratio is $ \text{C}_4\text{H}_{10} : \text{O}_2 = 2 : 13 $.
Step 2: Calculate the Required Moles of ( $\text{O}_2 $)
For 0.20 mol of $ \text{C}_4\text{H}_{10} $:
$$
\text{Required moles of } \text{O}_2 = 0.20 \, \text{mol} \times \frac{13}{2} = 1.30 \, \text{mol}
$$
Step 3: Compare Available Moles
- You have 2.6 mol of $ \text{O}_2 $, which is more than the 1.30 mol required.
- Therefore, $ \text{C}_4\text{H}_{10} $ is the limiting reactant.
Step 4: Determine the Excess
The excess $ \text{O}_2 $ is:
$$
\text{Excess } \text{O}_2 = 2.6 \, \text{mol} - 1.30 \, \text{mol} = 1.30 \, \text{mol}
$$
Limiting Reactants and Chemical Reactions
Theoretical Yield: How Much Product Can Be Made?
Theoretical yield
The theoretical yield is the maximum amount of product that can be formed based on the limiting reactant. It assumes the reaction proceeds to completion with no side reactions or losses.
Calculating Theoretical Yield:
- Identify the limiting reactant.
- Use the balanced equation to find the mole ratio between the limiting reactant and the desired product.
- Convert the moles of the product into mass, volume, or another unit if required.
Calculating Theoretical Yield
Using the same reaction as in the previous example, calculate the theoretical yield of $ \text{CO}_2 $ if 0.20 mol of $ \text{C}_4\text{H}_{10} $ is the limiting reactant.
Solution
Step 1: Use the Mole Ratio
From the balanced equation:
$$
2 \text{C}_4\text{H}_{10} \rightarrow 8\text{CO}_2
$$
The mole ratio is $ \text{C}_4\text{H}_{10} : \text{CO}_2 = 2 : 8 $, or $ 1 : 4 $.
Step 2: Calculate Moles of $ \text{CO}_2 $
For 0.20 mol of $ \text{C}_4\text{H}_{10} $:
$$
\text{Moles of } \text{CO}_2 = 0.20 \, \text{mol} \times 4 = 0.80 \, \text{mol}
$$
Step 3: Convert to Mass (Optional)
The molar mass of $ \text{CO}_2 $ is $ 44.01 \, \text{g mol}^{-1} $. Therefore:
$$
\text{Mass of } \text{CO}_2 = 0.80 \, \text{mol} \times 44.01 \, \text{g mol}^{-1} = 35.21 \, \text{g}
$$
- Many students forget to use the limiting reactant when calculating the theoretical yield.
- Always confirm which reactant limits the reaction first!
- What is the definition of a limiting reactant? Why is it important to identify the limiting reactant before calculating the theoretical yield?
- A reaction requires 3 moles of $ \text{A} $ to react with 5 moles of $ \text{B} $. If you start with 6 moles of $ \text{A} $ and 8 moles of $ \text{B} $, which reactant is limiting?
- Calculate the theoretical yield of $ \text{C} $ if the reaction $ \text{A} + 2\text{B} \rightarrow \text{C} $ starts with 2.0 mol of $ \text{A} $ and 4.0 mol of $ \text{B} $.
