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Preparation of Alkyl Halides

2 mol of Fe...

Question

2 mol of $F e S O_{4}$ in acid medium is oxidized by $x$ mol of $K M n O_{4}$ , whereas 2 mol of $F e C_{2} O_{4}$ in acid medium is oxidized by $y$ mol of $K M n O_{4}$ . The ratio of $x$ and $y$ is :

A

\frac{1}{3}

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B

\frac{1}{2}

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C

\frac{1}{4}

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D

\frac{1}{5}

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Solution

The correct option is A $\frac{1}{3}$
The balanced reactions are given below:
$2 K M n O_{4} + 10 F e S O_{4} + 8 H_{2} S O_{4} \to 5 F e_{2} (S O_{4})_{3} + K_{2} S O_{4} + 2 M n S O_{4} + 8 H_{2} O$
and
$10 F e C_{2} O_{4} + 6 K M n O_{4} + 24 H_{2} S O_{4} \to 3 K_{2} S O_{4} + 6 M n S O_{4} + 5 F e_{2} (S O_{4})_{3} + 24 H_{2} O + 20 C O_{2}$
So, from the reactions, we get
$\frac{x}{y} = \frac{4}{10} \times \frac{10}{12} = \frac{1}{3}$ .

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