The number of moles $H_{2} O_{2}$ needed to reduce 1 mole of $K M n O_{4}$ in acidic medium is

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2.5

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Solution

The correct option is B 2.5
$2 K M n O_{4} + 3 H_{2} S O_{4} + 5 H_{2} O_{2} \to 2 M n S O_{4} + K_{2} S O_{4} + 8 H_{2} O + 5 O_{2}$
1 mole
$2 K M n O_{4} = 5 H_{2} O_{2}$
$1 K M n O_{4} = 2.5 H_{2} O_{2}$

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