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Mole Concept

0.804 g sampl...

Question

$0.804 g$ sample of iron ore was dissolved in acid. Iron was reduced to $+ 2$ state and it required $47.2 m L$ of $0.112 N K M n O_{4}$ solution for titration. Calculate the percentage of iron of $F e_{3} O_{4}$ in the ore.

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Solution

The titration involves the conversion of ferrou into ferric,
$5 F e^{2 +} + M n O_{4}^{-} + 8 H^{+} \to 5 F e^{3 +} + M n^{2 +} + 4 H_{2} O$
$47.2 m L$ of $0.112 N K M n O_{4} \equiv 47.2 m L$ of $0.112 N F e^{2 +}$ ions
$= \frac{57.2 \times 0.112 \times 55.5}{1000} = 0.2934$
Mass of iron $= 0.2934 g$
% of iron in the ore $= \frac{0.2934}{0.804} \times 100 = 36.49$
$3 F e (3 \times 55.5) \to F e_{3} O_{4} (3 \times 55.5 + 64)$
$3 \times 55.5 g$ of iron form $230.5 g$ of $F e_{3} O_{4}$ .
$0.2934 g$ of iron will form $= \frac{230.5}{166.5} \times 0.2934 = 0.406 g$

% of $F e_{3} O_{4}$ in the ore $= \frac{0.406}{0.804} \times 100 = 50.5$ .

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