Finding the Mass of the Excess Reactant

Process Overview

1. Identify the Limiting Reactant
2. Calculate Grams of Excess Reactant Consumed
3. Determine Mass of Excess Reactant Remaining

Example: Aluminum and Sulfuric Acid Reaction

• Reaction: Aluminum reacts with sulfuric acid to produce hydrogen gas and aluminum sulfate
• Balanced Equation
  • Aluminum: Al
  • Sulfuric Acid: H2SO4
  • Aluminum Sulfate: Al2(SO4)3
  • Steps to balance:
    • 3 H2SO4 (for SO4 balance)
    • 2 Al (for Al balance)
    • 3 H2 (for H balance)
• Calculating Limiting Reactant
  • 80g Aluminum, 40g Sulfuric Acid
  • Molar Mass Calculation:
    • Al: 26.98 g/mol
    • H2SO4: 98.086 g/mol
  • Moles Calculation:
    • 80g Al yields 2.96 moles
    • 40g H2SO4 yields 0.4078 moles
  • Limiting Reactant: H2SO4 (lower mole/coefficient ratio)
• Excess Reactant Consumed
  • Convert grams of limiting reactant to grams of excess reactant
  • 7.33g of Aluminum is consumed
• Mass of Excess Reactant Remaining
  • Start: 80g of Aluminum
  • Used: 7.33g
  • Remaining: 72.67g

Practice Problem: Aluminum and Sulfur Reaction

• Reaction: Aluminum reacts with sulfur to produce aluminum sulfide
• Balanced Equation
  • Aluminum Sulfide: Al2S3
  • Balance using:
    • 3 S8 and 8 Al2S3
    • 16 Al (for Al balance)
• Calculating Limiting Reactant
  • 50g Aluminum, 70g Sulfur
  • Molar Mass Calculation:
    • Al: 26.98 g/mol
    • S8: 256.56 g/mol
  • Moles Calculation:
    • 50g Al yields 1.853 moles
    • 70g S8 yields 0.2728 moles
  • Limiting Reactant: S8 (lower mole/coefficient ratio)
• Excess Reactant Consumed
  • Convert grams of limiting reactant to grams of excess reactant
  • 39.26g of Aluminum is consumed
• Mass of Excess Reactant Remaining
  • Start: 50g of Aluminum
  • Used: 39.26g
  • Remaining: 10.74g

Conclusion

• Use mole/coefficient ratios to determine limiting reactants
• Always convert from the limiting reactant to determine excess reactant consumption
• Correct subtraction yields the remaining excess reactant