Limiting Reactant Practice Problems

Limiting Reactants and Stoichiometry

Overview

  • Focus on identifying limiting reactants.

  • Solve stoichiometry problems involving limiting reactants.

Reaction Example: Zinc and Hydrochloric Acid

  1. Balanced Chemical Reaction

    • Reactants: Zinc (Zn) + Hydrochloric Acid (HCl)

    • Products: Hydrogen Gas (H₂) + Zinc Chloride (ZnCl₂)

    • Balanced Equation:

      • Zn + 2 HCl → H₂ + ZnCl₂

  2. Identifying Limiting Reactant

    • Part A:

      • Given: 12 atoms of zinc, 8 molecules of HCl.

      • Calculation:

        • Zn: 12 particles (coefficient = 1) → 12/1 = 12

        • HCl: 8 particles (coefficient = 2) → 8/2 = 4

      • Conclusion: HCl is the limiting reactant because it has the lowest quantity per coefficient ratio.

    • Part B:

      • Given: 3 moles of Zinc and 4 moles of HCl.

      • Calculation:

        • Zn: 3 moles (coefficient = 1) → 3/1 = 3

        • HCl: 4 moles (coefficient = 2) → 4/2 = 2

      • Conclusion: HCl is again the limiting reactant even with more moles of Zn present.

    • Part C:

      • Given: Mass of Zinc (40 g) and HCl (56 g).

      • Convert grams to moles:

        • Molar mass of Zn = 65.39 g/mol

        • Molar mass of HCl = 36.458 g/mol

        • Moles of Zn: 40 g / 65.39 g/mol ≈ 0.6117 moles

        • Moles of HCl: 56 g / 36.458 g/mol ≈ 1.536 moles

      • Calculation:

        • Zn: 0.6117 moles (coef = 1) → 0.6117

        • HCl: 1.536 moles (coef = 2) → 1.536/2 ≈ 0.768

      • Conclusion: Zn is the limiting reactant due to the lower quantity.

Reaction Example: Ethane and Oxygen

  1. Balanced Chemical Reaction

    • Reactants: Ethane (C₂H₆) + Oxygen (O₂)

    • Products: Carbon Dioxide (CO₂) + Water (H₂O)

    • Balanced Equation:

      • 2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O

  2. Identifying Limiting Reactant

    • Part A:

      • Given: 5 moles of ethane, 16 moles of O₂.

      • Calculation:

        • Ethane: 5 moles (coef = 2) → 5/2 = 2.5

        • O₂: 16 moles (coef = 7) → 16/7 ≈ 2.29

      • Conclusion: O₂ is the limiting reactant.

    • Product Calculation:

      • From Ethane: 5 moles C₂H₆ → 10 moles CO₂ produced.

      • From Oxygen: 16 moles O₂ → (16 × 4) / 7 ≈ 9.14 moles CO₂ produced.

      • Theoretical Yield: 9.14 moles (less product determines the yield).

    • Part B:

      • Given: 30 g of ethane, 84 g of O₂.

      • Convert grams to moles:

        • Molar mass of C₂H₆ ≈ 30.068 g/mol

        • Molar mass of O₂ = 32 g/mol

      • Calculation:

        • Ethane: 30 g / 30.068 g/mol ≈ 0.999 moles

        • O₂: 84 g / 32 g/mol = 2.625 moles

        • Conversion to Grams of Water:

          • Ethane to H₂O: (30 / 30.068) * (6/2) * 18.016 = 53.93 g

          • O₂ to H₂O: (84 / 32) * (6/7) * 18.016 ≈ 40.54 g

      • Conclusion: 40.54 g is the lower theoretical yield, making O₂ the limiting reactant.

Review of Key Concepts

  • Limiting Reactant: Reactant that runs out first and stops the reaction.

  • Stoichiometry: Calculation of reactants and products in chemical reactions should involve balanced equations and mole ratios.

  • Theoretical Yield: Maximum amount of product obtainable in a chemical reaction, based on the limiting reactant.