Core Concept

In a chemical reaction, the limiting reactant (or limiting reagent) is the substance that is completely used up first, stopping the reaction from continuing and thus determining the maximum amount of product that can be formed. The excess reactant is the reactant that remains after the limiting reactant is fully consumed.

Key Concepts

1. Limiting Reactant:

   • The reactant that is entirely consumed in a reaction, limiting the formation of products.

   • Determines the maximum amount of product that can be formed.

2. Excess Reactant:

   • The reactant that remains after the limiting reactant is used up.

   • Does not affect the amount of product formed but can be calculated to determine the leftover quantity.

3. Significance in Stoichiometry:

   • Identifying the limiting reactant allows you to accurately calculate the theoretical yield of the reaction, which is the maximum amount of product possible based on the limiting reactant.

Steps to Identify the Limiting Reactant

1. Write and Balance the Chemical Equation:

   • Ensure the equation is balanced with the correct coefficients.

2. Convert Masses to Moles (if necessary):

   • Convert the given amounts of reactants (often in grams) to moles using their molar masses.

3. Use Mole Ratios to Compare Reactants:

   • Use the mole ratios from the balanced equation to determine the amount of product each reactant could theoretically produce.

   • The reactant that produces the smallest amount of product is the limiting reactant.

4. Calculate the Amount of Product:

   • Use the limiting reactant to determine the theoretical yield of the reaction.

5. Calculate Excess Reactant (if needed):

   • If required, calculate how much of the excess reactant remains after the reaction.

Example Problem: Identifying the Limiting Reactant

Problem: Given 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 limiting reactant and calculate the amount of water (H2O\text{H}_2\text{O}H2​O) produced.

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 already balanced.

2. Convert Masses 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. Use Mole Ratios to Compare Reactants:

   • From 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 would 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 the available 0.625 mol O20.625 \, \text{mol O}_20.625mol O2​. Therefore, O2\text{O}_2O2​ is the limiting reactant.

4. Calculate the Amount of Product:

   • Using O2\text{O}_2O2​ as the limiting reactant: 0.625 mol O2×2 mol H2O1 mol O2=1.25 mol H2O0.625 \, \text{mol O}_2 \times \frac{2 \, \text{mol H}_2\text{O}}{1 \, \text{mol O}_2} = 1.25 \, \text{mol H}_2\text{O}0.625mol O2​×1mol O2​2mol H2​O​=1.25mol H2​O

   • Molar mass of H2O=18.02 g/mol\text{H}_2\text{O} = 18.02 \, \text{g/mol}H2​O=18.02g/mol: 1.25 mol H2O×18.02 g/mol=22.53 g H2O1.25 \, \text{mol H}_2\text{O} \times 18.02 \, \text{g/mol} = 22.53 \, \text{g H}_2\text{O}1.25mol H2​O×18.02g/mol=22.53g H2​O

Answer: The limiting reactant is O2\text{O}_2O2​, and the maximum amount of H2O\text{H}_2\text{O}H2​O produced is 22.53 g.

Key Tips for Limiting Reactant Problems

• Always Balance the Equation First: Mole ratios are essential for identifying the limiting reactant.

• Convert to Moles First: Always convert masses to moles before using mole ratios.

• Compare Based on Products: Calculate the amount of product each reactant can form, and the smaller amount indicates the limiting reactant.

• Double-Check Units: Be mindful of units (grams, moles) to avoid calculation errors.