The volume, in mL, of 0.02 M K Cr O solution required to react with 0.288 g of ferrous oxalate in ac — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

**Step 1: Write the balanced equation** In acidic medium, potassium dichromate oxidizes ferrous oxalate: 2K₂Cr₂O₇ + 6FeC₂O₄ + 8H₂SO₄ → 2Cr₂(SO₄)₃ + 2K₂SO₄ + 6FeSO₄ + 2CO₂ + 8H₂O Simplified: 2Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O **Step 2: Calculate molar mass of FeC₂O₄** M(FeC₂O₄) = 56 + 2(12) + 4(16) = 56 + 24 + 64 = 144 g/mol **Step 3: Calculate moles of ferrous oxalate** n(FeC₂O₄) = 0.288 g ÷ 144 g/mol = 0.002 mol **Step 4: Determine moles of K₂Cr₂O₇ needed** From the balanced equation: 6 mol FeC₂O₄ reacts with 2 mol Cr₂O₇²⁻ Mole ratio = 2:6 = 1:3 n(K₂Cr₂O₇) = 0.002 mol × (2/6) = 0.002 × (1/3) = 1/500 mol = 0.00033... mol Wait, recalculating: 0.002 × (2/6) = 0.002/3 ≈ 6.67 × 10⁻⁴ mol Actually: n = 0.002 × (2/6) = 0.0004/3... Let me recalculate: 0.002/3 =