The weight of $K M n O_{4}$ required to completely oxidise $0.25$ moles of $F e S O_{4}$ in acidic medium is:

[Molecular weight of $K M n O_{4} = 158$ ]

5.8

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1.5

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7.9

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0.79

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Solution

The correct option is C $7.9$
The balanced chemical equation is as follows:
$2 K M n O_{4} + 3 H_{2} S O_{4} + 10 F e S O_{4} \to K_{2} S O_{4} + 2 M n S O_{4} + 5 F e_{2} (S O_{4})_{3} + 3 H_{2} O$ .

2 moles $(2 \times 158 g)$ of $K M n O_{4}$ completely oxidizes $10$ moles of $F e S O_{4}$ .

The weight of $K M n O_{4}$ required to completely oxidize $0.25$ moles of $F e S O_{4}$ will be $\frac{(2 \times 158)}{10} \times 0.25 = 7.9 g$ .

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