A 0.2 g sample of chromite was fused with excess of Na₂O₂ and brought into solution according to reaction: 2Fe(CrO₂)₂ + 7Na₂O₂ → 2NaFeO₂ + 4Na₂CrO₄ + 2Na₂O. The solution was acidified with dil. HCl an

Question

A 0.2 g sample of chromite was fused with excess of Na₂O₂ and brought into solution according to reaction: 2Fe(CrO₂)₂ + 7Na₂O₂ → 2NaFeO₂ + 4Na₂CrO₄ + 2Na₂O. The solution was acidified with dil. HCl and 1.96 g Mohr's salt (molar mass = 392 g/mol) was added. The excess of Fe²⁺ required 40 ml of 0.05 N - K₂Cr₂O₇ for titration. What is the per cent of Cr in sample? (Cr = 52, Fe = 56)

TestHub · Accepted Answer

Mohr's salt = 1.96/392 = 5 mmol Fe²⁺ = 5 meq. Excess Fe²⁺ = 40×0.05 = 2 meq. Fe²⁺ consumed by chromate = 3 meq = equivalents of Cr. Each Cr goes +3 → +6 (3e⁻), so eq wt of Cr = 52/3. Mass Cr = 3×52/3 = 52 mg = 0.052 g. %Cr = 0.052/0.2×100 = 26%. Ans 0026.

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