Core Concept

the excess reactant (or excess reagent) is the reactant that remains after the limiting reactant is completely used up. The excess reactant is not entirely consumed, meaning some amount of it will be left over once the reaction stops. Understanding how to identify and calculate the leftover amount of the excess reactant is important for accurately predicting the reaction’s outcome and for practical applications like reducing waste.

Key Concepts

1. Excess Reactant:

   • The reactant that is not fully consumed in a chemical reaction.

   • It is present in a larger quantity than needed to react completely with the limiting reactant.

2. Limiting Reactant:

   • The reactant that determines the maximum amount of product formed.

   • It is completely consumed in the reaction, causing the reaction to stop.

3. Importance of Excess Reactant Calculation:

   • Helps understand reaction efficiency and minimize waste.

   • Useful for calculating the actual yield of a reaction in practical applications.

Steps to Determine the Amount of Excess Reactant Left Over

1. Write and Balance the Chemical Equation:

   • Ensure the reaction equation is balanced, which allows for accurate mole ratios between reactants.

2. Identify the Limiting and Excess Reactants:

   • Convert the given amounts of reactants (usually in grams) to moles.

   • Use mole ratios from the balanced equation to determine which reactant will be completely used up first (the limiting reactant) and which will remain (the excess reactant).

3. Calculate Moles of Excess Reactant Used:

   • Use the limiting reactant to determine how much of the excess reactant is required to react with it.

   • Apply mole ratios to calculate the moles of the excess reactant that actually participate in the reaction.

4. Find the Remaining Amount of Excess Reactant:

   • Subtract the moles of excess reactant used from the total moles of excess reactant initially present.

   • Convert the leftover moles of excess reactant back to grams (if needed) by multiplying by its molar mass.

Example Problem: Finding the Excess Reactant Left Over

Problem: For the reaction below, if 5.0 g5.0 \, \text{g}5.0g of H2\text{H}_2H2​ reacts with 20.0 g20.0 \, \text{g}20.0g of O2\text{O}_2O2​, identify the excess reactant and calculate how much of it is left over after the reaction.

2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}2H2​+O2​→2H2​O

Solution:

1. Write and Balance the Equation:

   • The equation is balanced as written.

2. Convert Given Amounts to Moles:

   • Molar mass of H2=2.02 g/mol\text{H}_2 = 2.02 \, \text{g/mol}H2​=2.02g/mol: Moles of H2=5.0 g2.02 g/mol=2.48 mol\text{Moles of } \text{H}_2 = \frac{5.0 \, \text{g}}{2.02 \, \text{g/mol}} = 2.48 \, \text{mol}Moles of H2​=2.02g/mol5.0g​=2.48mol

   • Molar mass of O2=32.00 g/mol\text{O}_2 = 32.00 \, \text{g/mol}O2​=32.00g/mol: Moles of O2=20.0 g32.00 g/mol=0.625 mol\text{Moles of } \text{O}_2 = \frac{20.0 \, \text{g}}{32.00 \, \text{g/mol}} = 0.625 \, \text{mol}Moles of O2​=32.00g/mol20.0g​=0.625mol

3. Identify the Limiting and Excess Reactants:

   • According to the balanced equation, 2 mol H22 \, \text{mol H}_22mol H2​ reacts with 1 mol O21 \, \text{mol O}_21mol O2​.

   • To react with 0.625 mol O20.625 \, \text{mol O}_20.625mol O2​, we need: 0.625 mol O2×2 mol H21 mol O2=1.25 mol H20.625 \, \text{mol O}_2 \times \frac{2 \, \text{mol H}_2}{1 \, \text{mol O}_2} = 1.25 \, \text{mol H}_20.625mol O2​×1mol O2​2mol H2​​=1.25mol H2​

   • We have 2.48 mol H22.48 \, \text{mol H}_22.48mol H2​, which is more than enough to react with 0.625 mol O20.625 \, \text{mol O}_20.625mol O2​. Therefore, O2\text{O}_2O2​ is the limiting reactant, and H2\text{H}_2H2​ is the excess reactant.

4. Calculate Moles of Excess Reactant Used:

   • Using 0.625 mol O20.625 \, \text{mol O}_20.625mol O2​ (the limiting reactant), we find the moles of H2\text{H}_2H2​ used: 0.625 mol O2×2 mol H21 mol O2=1.25 mol H20.625 \, \text{mol O}_2 \times \frac{2 \, \text{mol H}_2}{1 \, \text{mol O}_2} = 1.25 \, \text{mol H}_20.625mol O2​×1mol O2​2mol H2​​=1.25mol H2​

5. Find the Remaining Amount of Excess Reactant:

   • Subtract the moles of H2\text{H}_2H2​ used from the initial moles of H2\text{H}_2H2​: 2.48 mol H2−1.25 mol H2=1.23 mol H22.48 \, \text{mol H}_2 - 1.25 \, \text{mol H}_2 = 1.23 \, \text{mol H}_22.48mol H2​−1.25mol H2​=1.23mol H2​

   • Convert moles of remaining H2\text{H}_2H2​ back to grams: 1.23 mol H2×2.02 g/mol=2.49 g H21.23 \, \text{mol H}_2 \times 2.02 \, \text{g/mol} = 2.49 \, \text{g H}_21.23mol H2​×2.02g/mol=2.49g H2​

Answer: After the reaction, 2.49 g of H2\text{H}_2H2​ remains as the excess reactant.

Key Tips for Excess Reactant Calculations

• Balance the Equation First: Accurate mole ratios are necessary to identify the limiting and excess reactants.

• Use the Limiting Reactant to Calculate Excess Used: Once the limiting reactant is identified, use it to find how much of the excess reactant is actually consumed.

• Double-Check Units: Keep track of units (grams, moles) to avoid errors in conversions.

• Practice with Different Reactions: Excess reactant calculations are useful in various contexts, so practicing with diverse problems helps reinforce the process.