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)
How many millimoles of Prussian blue is formed?
Options:
Answer:
Solution:
K₄Fe(CN)₆ taken = 10 ml × 1 M = 10 mmol. Excess ferrocyanide precipitated as K₂Fe[Fe(CN)₆] (1 Fe²⁺ : 1 ferrocyanide): 6 ml × 0.4 N Fe²⁺ (n = 2, so 0.2 M) = 1.2 mmol. Fe³⁺ consumed = 10 - 1.2 = 8.8 mmol, and KFe[Fe(CN)₆] forms 1:1 with Fe³⁺, so Prussian blue = 8.8 mmol, matching (b).
