Chemical Calculations — A/L Chemistry

How do I practise Chemical Calculations?

1 question with a worked solution

Check before you answer: Mole Ratios & Limiting Reagents: Stepwise calculation identifying the limiting reactant and theoretical yield using balanced chemical equations.

1. A 25.00 cm³ sample of 0.100 mol dm⁻³ ethanedioic acid ((COOH)₂) requires 20.00 cm³ of acidified KMnO₄ solution for complete reaction. What is the molar concentration of the KMnO₄ solution?
   1. A) 0.0400 mol dm⁻³
   2. B) 0.0500 mol dm⁻³
   3. C) 0.0200 mol dm⁻³
   4. D) 0.1250 mol dm⁻³
   Check my answer

   Correct answer: B) 0.0500 mol dm⁻³

   The balanced redox equation is 2MnO₄⁻ + 5(COOH)₂ + 6H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O. The mole ratio is 2 moles of KMnO₄ to 5 moles of oxalic acid. Moles of oxalic acid = 0.02500 × 0.100 = 0.002500 mol. Moles of KMnO₄ = (2/5) × 0.002500 = 0.001000 mol. Concentration = 0.001000 / 0.02000 = 0.0500 mol dm⁻³.

   Show worked solution

   Step 1: Write half reactions and overall balanced equation: 5 × [(COOH)₂ → 2CO₂ + 2H⁺ + 2e⁻] 2 × [MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O] Overall: 2MnO₄⁻ + 5(COOH)₂ + 6H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O Step 2: Calculate moles of (COOH)₂: n₁ = C₁ × V₁ = 0.100 mol dm⁻³ × (25.00 / 1000) dm³ = 2.50 × 10⁻³ mol. Step 3: Calculate moles of KMnO₄: n₂ = (2/5) × n₁ = (2/5) × 2.50 × 10⁻³ mol = 1.00 × 10⁻³ mol. Step 4: Calculate molarity of KMnO₄: C₂ = n₂ / V₂ = (1.00 × 10⁻³ mol) / (20.00 / 1000 dm³) = 0.0500 mol dm⁻³. Answer: 0.0500 mol dm⁻³.
   Modelled on: the A/L Chemistry examination pattern

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Where does Chemical Calculations fit in the syllabus?

Grade 12 · Term 1