How many $m o l e s$ of acidified $F e S O_{4}$ can be completely oxidised by $o n e m o l e$ of $K M n O_{4}$ ?

10

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5

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6

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2

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Solution

The correct option is B $5$
$2 K M n O_{4} + 3 H_{2} S O_{4} \to K_{2} S O_{4} + 2 M n S O_{4} + 3 H_{2} O + 5 [O]$
$[2 F e S O_{4} + H_{2} S O_{4} + [O] \to F e_{2} (S O_{4})_{3} + H_{2} O] \times 5 - ----------------------------------------------------- -$
$2 K M n O_{4} + 10 F e S O_{4} + 8 H_{2} S O_{4} \to K_{2} S O_{4} + 2 M n S O_{4} + 5 F e_{2} (S O_{4})_{3} + 8 H_{2} O - ----------------------------------------------------------------------------------- -$
$2 m o l e K M n O_{4} \equiv 10 m o l e F e S O_{4}$
$1 m o l e K M n O_{4} \equiv 5 m o l e F e S O_{4}$ .

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