The volume, in mL, of 0.02 M K₂Cr₂O₇ solution required to react with 0.288 g of ferrous oxalate in acidic medium is _______.
(Molar mass of Fe = 56 g mol –1 )

Solution

Chemistry ma’am · 4 steps

Ferrous oxalate loses electrons twice, from iron and from oxalate, so its n-factor is 3, not 1.

1. Moles of salt. Ferrous oxalate is FeC₂O₄, with molar mass 56 + 24 + 64 = 144 g mol⁻¹. So 0.288 g is 0.288/144 = 0.002 mol.
2. Two reducing parts. Dichromate oxidises both ions: Fe²⁺ → Fe³⁺ gives 1 electron, and C₂O₄²⁻ → 2CO₂ gives 2 electrons. Each FeC₂O₄ gives 3 electrons, while each Cr₂O₇²⁻ takes 6 electrons on going to 2Cr³⁺.

   0.002 × 3 = 0.02 × 6 × V

3. Solve for V. V = 0.006/0.12 = 0.05 L. That is 0.001 mol of K₂Cr₂O₇.
4. So. The volume needed is 0.05 L, which is 50 mL.

The common slip: 16.67. Counting only Fe²⁺ → Fe³⁺ (n-factor 1) and ignoring the oxalate gives about 16.67 mL, not 50.

The idea to keep. When a salt has a reducing cation and a reducing anion, add the electrons lost by both to get its n-factor. Then electrons lost by the reducer equal electrons gained by the oxidiser.

Answer from the NTA final answer key. Solution by PrepWiser.