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Quantitative Chemistry – Lecture 10: Worked Examples Part 2 – Advanced Multi-Step Problems Including Reactions in Series, Yields, and Industrial Applications

Quantitative Chemistry – Lecture 10: Worked Examples Part 2 – Advanced Multi-Step Problems Including Reactions in Series, Yields, and Industrial Applications

Introduction

Lecture 10 focuses on advanced quantitative chemistry problems, integrating concepts from Lectures 1-9. This session emphasises:

  • Multi-step reactions
  • Calculating theoretical and actual yields
  • Percent yield and efficiency
  • Industrial and laboratory applications

By the end of this lecture, students will be able to solve complex real-world chemistry problems, applying moles, mass, concentration, and stoichiometry in sequence.

1. Multi-Step Reaction Example

Problem: Ammonia is produced via the Haber process:

Calculate the mass of NH₃ produced from 28 g N₂ reacting with 6 g H₂.

Step 1: Moles of reactants:

  • N₂: 28 / 28 = 1 mol
  • H₂: 6 / 2.016 ≈ 2.98 mol

Step 2: Determine limiting reactant:

  • Required H₂ for 1 mol N₂ = 3 mol → only 2.98 mol available → H₂ limiting

Step 3: Moles of NH₃:

  • H₂ reacts in 3:2 ratio → 2.98 × (2/3) ≈ 1.987 mol NH₃

Step 4: Mass of NH₃:

  • Molar mass NH₃ = 14 + 3 × 1 = 17 g/mol
  • Mass = 1.987 × 17 ≈ 33.78 g

Answer: 33.78 g NH₃

2. Percent Yield Calculation

Problem: The theoretical yield of sodium chloride from 5 g Na reacting with Cl₂ is 8.1 g. If the actual yield is 7.5 g, calculate the percentage yield.

Answer: 92.6% yield.

3. Reaction in Series

Problem: Magnesium reacts with HCl to produce MgCl₂ and H₂. H₂ then reacts with O₂ to produce H₂O. Determine the total mass of H₂O formed from 24 g Mg.

Step 1: Mg + 2 HCl → MgCl₂ + H₂

  • Moles Mg = 24 / 24.31 ≈ 0.987 mol
  • Moles H₂ produced = 0.987 mol (1:1 ratio)

Step 2: 2 H₂ + O₂ → 2 H₂O

  • Moles H₂O = 0.987 × (2/2) = 0.987 mol
  • Mass H₂O = 0.987 × 18.016 ≈ 17.78 g

Answer: 17.78 g H₂O

4. Limiting Reactant in Series

Problem: 10 g of Al reacts with 10 g Fe₂O₃:

Determine the limiting reactant and the mass of Fe produced.

Step 1: Moles:

  • Al = 10 / 26.98 ≈ 0.371 mol
  • Fe₂O₃ = 10 / 159.7 ≈ 0.0626 mol

Step 2: Stoichiometry: 2 Al : 1 Fe₂O₃ → 0.371 mol Al requires 0.371/2 ≈ 0.1855 mol Fe₂O₃

  • Only 0.0626 mol Fe₂O₃ → Fe₂O₃ limiting

Step 3: Moles Fe produced: 1 Fe₂O₃ → 2 Fe → 0.0626 × 2 = 0.1252 mol Fe

Step 4: Mass Fe = 0.1252 × 55.85 ≈ 6.99 g

Answer: 6.99 g Fe produced, Fe₂O₃ limiting.

5. Solution Preparation and Concentration

Problem: Prepare 250 mL of 0.5 M K₂SO₄ solution. How much salt is required?

Step 1: Moles required = M × V = 0.5 × 0.25 = 0.125 mol

Step 2: Molar mass K₂SO₄ = 2 × 39.10 + 32.07 + 4 × 16 = 174.27 g/mol

Step 3: Mass = 0.125 × 174.27 ≈ 21.78 g

Answer: Dissolve 21.78 g K₂SO₄ in 250 mL of water.

6. Titration Problem

Problem: 25.0 mL of H₂SO₄ reacts with 30.0 mL of 0.1 M NaOH. Determine the concentration of H₂SO₄.

Step 1: Moles NaOH = 0.1 × 0.030 = 0.003 mol

Step 2: Moles H₂SO₄ = 0.003 × (1/2) = 0.0015 mol

Step 3: Concentration H₂SO₄ = 0.0015 / 0.025 = 0.06 M

Answer: 0.06 M H₂SO₄

7. Industrial Chemistry Problem

Problem: Ammonium nitrate (NH₄NO₃) is produced from NH₃ and HNO₃. Calculate the mass of NH₄NO₃ from 17 g NH₃.

Step 1: Moles NH₃ = 17 / 17 = 1 mol

Step 2: Reaction ratio = 1:1 → 1 mol NH₄NO₃

Step 3: Mass NH₄NO₃ = 14 + 4 + 14 + 48 = 80 g/mol

Step 4: Mass = 1 × 80 = 80 g

Answer: 80 g NH₄NO₃

8. Yield in Industrial Context

Problem: Industrial reaction of 100 g N₂ with excess H₂ theoretically produces 117 g NH₃. Actual yield = 105 g. Calculate percent yield.

Answer: 89.7% yield

9. Series Reaction with Limiting Reagent

Problem: 20 g Al reacts with 50 g CuO:

Step 1: Moles:

  • Al = 20 / 26.98 ≈ 0.741 mol
  • CuO = 50 / 79.55 ≈ 0.629 mol

Step 2: Stoichiometry: 2:3 → Al required for 0.629 mol CuO = 0.629 × 2/3 ≈ 0.419 mol

  • Available Al = 0.741 mol → CuO limiting

Step 3: Moles Cu produced = 0.629 × 3 = 1.887 mol

Step 4: Mass Cu = 1.887 × 63.55 ≈ 119.9 g

Answer: 119.9 g Cu produced, CuO limiting.

10. Multi-Step Concentration Problem

Problem: 50 mL 0.2 M HCl is diluted to 200 mL. Calculate the new concentration.

Step 1: Use C₁V₁ = C₂V₂ → 0.2 × 0.05 = C₂ × 0.2

Step 2: C₂ = 0.01 / 0.2 = 0.05 M

Answer: 0.05 M HCl

11. Summary of Advanced Worked Examples

Lecture 10 illustrates:

  • Multi-step stoichiometry and limiting reagent identification
  • Percent yield and theoretical vs actual yield
  • Industrial reaction calculations
  • Solution preparation and dilution
  • Titration and concentration applications

These examples consolidate all prior lectures, linking theory to practical, laboratory, and industrial chemistry.

Further Reading:

This concludes the 10-lecture quantitative chemistry series, covering theory, calculations, and extensive worked examples.

Available in PDF format with an associated quiz

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