A sample of 1.0 g of solid Fe 2 O 3 of 80% purity is dissolved in a moderately concentrated HCl solution which is reduced by zinc dust. The resulting solution required 16.7 ml of a 0.1 M solution

Question

A sample of 1.0 g of solid Fe 2 ​ O 3 ​ of 80% purity is dissolved in a moderately concentrated HCl solution which is reduced by zinc dust. The resulting solution required 16.7 ml of a 0.1 M solution of the oxidant. Calculate the number of electrons taken up by the oxidant (the atomic weights of Fe and O are 56 and 16 respectively)

TestHub · Accepted Answer

Weight of pure Fe₂O₃ = 1 × 100 80 ​ = 0.8 g Fe₂O₃ + 6HCl ⟶ 2FeCl₃ + 3H₂O FeCl₃ + H 2 &#x2192; Z n &#xA0; d u s t FeCl₂ + HCl Fe²⁺ + oxidant ⟶ Fe³⁺ + reductant Equivalent weight of Fe₂O₃ = Eq u i v a l e n tw e i g h t o f F e _ 2 O _ 3 = 2 m o l ec u l a r w e i g h t ​ = 2 160 ​ = 80 g Milliequivalents of Fe₂O₃ = Milliequivalents of oxidants. 80 0.8 ​ × 1 0 3 = 16.7 × 0.1 × x x = 6 where x is the number of electrons taken by the oxidant.

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