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Word Problems

How many mL...

Question

How many $m L$ of a $0.05 M K M n O_{4}$ solution are required to oxidise $2.0 g$ of $F e S O_{4}$ in a dilute solution (acidic)?

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Solution

$10 F e S O_{4} 10 \times 151.8 + 2 K M n O_{4} 2 \times 158 + 8 H_{2} S O_{4} \to K_{2} S O_{4} + 2 M n S O_{4} + 5 F e_{2} (S O_{4})_{3} + 8 H_{2} O$
$10 \times 151.8 g$ of $F e S O_{4}$ require $K M n O_{4} = 2 \times 158 g$
$2 g$ of $F e S O_{4}$ will require $K M n O_{4} = \frac{2 \times 158 \times 2}{10 \times 151.8} g$
Suppose, $V m L$ of $K M n O_{4}$ solution $(0.05 M)$ is required.
Amount of $K M n O_{4}$ in this solution $= \frac{158 \times 0.05}{1000} \times V$
Thus, $\frac{158 \times 0.05 \times V}{1000} = \frac{2 \times 158 \times 2}{10 \times 151.8}$
$V = 52.7 m L$ .

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