Lecture Notes on Limiting Reactants and Stoichiometry

Introduction

• Focus on limiting reactants and stoichiometry problems.
• Identify reactants and solve for limiting reactants.

Example 1: Zinc and Hydrochloric Acid Reaction

Balanced Chemical Reaction

• Reactants: Zinc (Zn) + Hydrochloric Acid (HCl)
• Products: Hydrogen Gas (H₂) + Zinc Chloride (ZnCl₂)
• Balanced Equation: [ ext{Zn} + 2 ext{HCl} ightarrow ext{H}_2 + ext{ZnCl}_2 ]

Identifying Limiting Reactants

Part A

• Given: 12 atoms of Zinc and 8 molecules of HCl.
• Calculation:
  • Zinc: [ 12 ext{ atoms} / 1 = 12 ]
  • HCl: [ 8 ext{ molecules} / 2 = 4 ]
• Limiting Reactant: HCl (lower quantity per coefficient ratio).

Part B

• Given: 3 moles of Zinc and 4 moles of HCl.
• Calculation:
  • Zinc: [ 3 ext{ moles} / 1 = 3 ]
  • HCl: [ 4 ext{ moles} / 2 = 2 ]
• Limiting Reactant: HCl (lower quantity per coefficient ratio).

Part C

• Given: 40 grams of Zinc and 56 grams of HCl.
• Convert grams to moles:
  • Molar mass of Zn = 65.39 g/mol
  • Molar mass of HCl = 36.458 g/mol
  • Moles of Zn: [ 40 / 65.39 = 0.6117 ]
  • Moles of HCl: [ 56 / 36.458 = 1.536 ]
• Ratio Calculation:
  • Zinc: [ 0.6117 / 1 = 0.6117 ]
  • HCl: [ 1.536 / 2 = 0.768 ]
• Limiting Reactant: Zinc (lower quantity per coefficient ratio).

Example 2: Ethane and Oxygen Reaction

Balanced Chemical Reaction

• Reactants: Ethane (C₂H₆) + Oxygen (O₂)
• Products: Carbon Dioxide (CO₂) + Water (H₂O)
• Balanced Equation: [ 2 ext{C}_2 ext{H}_6 + 7 ext{O}_2 ightarrow 4 ext{CO}_2 + 6 ext{H}2 ext{O} ]

Identifying Limiting Reactants

Part A

• Given: 5 moles of Ethane and 16 moles of Oxygen.
• Ratio Calculation:
  • Ethane: [ 5 / 2 = 2.5 ]
  • Oxygen: [ 16 / 7 \approx 2.29 ]
• Limiting Reactant: O₂ (lower quantity per coefficient ratio).

Calculating Moles of CO₂ Produced

• From Ethane:
  • [ 5 ext{ moles of C}_2 ext{H}_6 ightarrow (5 imes 4 / 2) = 10 ext{ moles CO}_2 ]
• From Oxygen:
  • [ 16 ext{ moles O}_2 ightarrow (16 imes 4 / 7) ext{ moles CO}_2 \approx 9.14 \text{ moles CO}_2 ]
• Theoretical Yield: 9.14 moles CO₂ (from limiting reactant O₂).

Part B

• Given: 30 grams of Ethane and 84 grams of O₂.
• Convert grams to moles and calculate product yield:
  • Molar mass of Ethane = 30.068 g/mol
  • Molar mass of O₂ = 32 g/mol
  • Moles to grams conversion for H₂O:
    • Ethane to H₂O: 53.93 grams
    • O₂ to H₂O: 40.54 grams
• Theoretical Yield of Water: 40.54 grams (from limiting reactant O₂).

Conclusion

• The limiting reactant determines the theoretical yield of products in a reaction.
• Always calculate the yield from both reactants and determine the lower value to find the limiting reactant.