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Oxidising Reagent and Reducing Reagent

A 3 g sample ...

Question

A 3 g sample of haematite (iron ore containing $F e_{2} O_{3}$ ) is dissolved in hydrochloric acid and the volume is made up to 250 mL. 25 mL of this solution was reduced with $S n C l_{2}$ . Unreacted $S n C l_{2}$ was removed by extraction, and the residual solution needed 30 mL of 0.02 M acidified $K_{2} C r_{2} O_{7}$ solution to reach equivalence point. Determine the mass percentage of $F e_{2} O_{3}$ in the ore.

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Solution

$F e_{2} O_{3} + H C l \to F e C l_{3} + H_{2} O . . . (1) F e C l_{3} + S n C l_{2} \to F e C l_{2} + S n C l_{4} . . . (2) F e^{2 +} + C r_{2} O_{7}^{2 -} \to F e^{3 +} + C r^{3 +} . . . (3)$
Note that (1) is NOT a redox reaction
On balancing:
$F e_{2} O_{3} + 6 H C l \to 2 F e C l_{3} + 3 H_{2} O . . . (4) 2 F e C l_{3} + S n C l_{2} \to 2 F e C l_{2} + S n C l_{4} . . . (5) F e^{2 +} + C r_{2} O_{7}^{2 -} \to F e^{3 +} + 2 C r^{3 +} . . . (6)$
From equation (4), moles of $F e_{2} O_{3} = 1 / 2$ moles of $F e^{3 +}$ ions.

Moles of $K_{2} C r_{2} O_{7} = 0.02 \times 30 / 1000 = 6 \times 10^{- 4}$
Gram Equivalents of $K_{2} C r_{2} O_{7} = 6 \times 6 \times 10^{- 4}$
⇒gram equivalents of $F e^{2 +} = 6 \times 6 \times 10^{- 4}$
⇒moles of $F e^{2 +} = 6 \times 6 \times 10^{- 4}$
⇒moles of $F e^{3 +} = 6 \times 6 \times 10^{- 4}$
⇒moles of $F e_{2} O_{3} = (6 \times 6 \times 10^{- 4}) / 2 = 3 \times 6 \times 10^{- 4} = 1.8 m m o l$
Molar mass of $F e_{2} O_{3} = 160 g,$
∴ mass of $F e_{2} O_{3} i n 25 m L = 1.8 \times 10^{- 3} \times 160 = 0.288 g$
Hence mass of $F e_{2} O_{3} i n 250 m L = 2.88 g$
Mass percentage $= \frac{2.88}{3} \times 100 = 96$ %

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