A 10 g mixture of Cu₂S and CuS was treated with 400 ml of 0.4 M - MnO₄⁻ in acid solution producing SO₂, Cu²⁺ and Mn²⁺. The SO₂ was boiled off and the excess of MnO₄⁻ was titrated with 200 ml of 1 M -

Question

A 10 g mixture of Cu₂S and CuS was treated with 400 ml of 0.4 M - MnO₄⁻ in acid solution producing SO₂, Cu²⁺ and Mn²⁺. The SO₂ was boiled off and the excess of MnO₄⁻ was titrated with 200 ml of 1 M - Fe²⁺ solution. The percentage of CuS in original mixture is (Cu = 64)

TestHub · Accepted Answer

MnO₄⁻ taken = 400×0.4 = 160 mmol. Excess: Fe²⁺ = 200×1 = 200 mmol = 200 meq, so MnO₄⁻ left = 200/5 = 40 mmol; consumed = 120 mmol = 600 meq. Let a, b = mol of Cu₂S (M = 160) and CuS (M = 96): 160a+96b = 10. Electrons: Cu₂S gives 8e⁻ (2Cu⁺ → 2Cu²⁺ + S → S⁴⁺), CuS gives 6e⁻: 8a+6b = 0.6. Solving, b = 1/12 mol, mass CuS = 96/12 = 8 g → 80%. Ans 0080.

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