Chemistry - Stoichiometry Question with Solution | TestHub
Chromium exists as FeCr₂O₄ in the nature and it contains Fe₀.₉₅O₁.₀₀ as an impurity. To obtain pure chromium from FeCr₂O₄, the ore is fused with KOH and oxygen is passed through the mixture when K₂CrO₄ and Fe₂O₃ are produced. A quantity of 2 g of ore required 280 ml of O₂ at 0°C and 1 atm for complete oxidation of ore. K₂CrO₄ is then precipitated as BaCrO₄ after addition of Barium salt. To the remaining solution, 10 ml of 1 M - K₄Fe(CN)₆ is added when Fe³⁺ ions reacts with it to form KFe[Fe(CN)₆], after called 'Prussian Blue'. To determine excess of K₄Fe(CN)₆ in solution, 6 ml of 0.4 N - Fe²⁺ is added when all the K₄Fe(CN)₆ is precipitated as K₂Fe[Fe(CN)₆]. (Fe = 56)
What is the percentage of Fe₀.₉₅O₁.₀₀ in the ore?
Options:
Answer:
Solution:
Let x g = FeCr₂O₄ (M = 224) and y g = Fe₀.₉₅O (M = 69.2). O₂ = 280/22400 = 12.5 mmol = 50 meq: 7x/224 + 0.85y/69.2 = 0.05 (i). Fe³⁺ formed reacts 1:1 with ferrocyanide; excess K₄Fe(CN)₆ = 6 × 0.4/2 = 1.2 mmol, so Fe³⁺ = 10 - 1.2 = 8.8 mmol: x/224 + 0.95y/69.2 = 0.0088 (ii). Solving (i) and (ii): y = 0.1384 g. Percentage = 0.1384/2 × 100 = 6.92%, matching (a).
