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

2

moles

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2.5

moles

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5

moles

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3

moles

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Solution

The correct option is D $2.5$ moles
The overall reaction is, $6 H^{+} + 2 M n O_{4}^{-} + 5 H_{2} O_{2} \to 5 O_{2} + 2 M n^{2 +} + 8 H_{2} O$ .

Therefore, the valency factor of $H_{2} O_{2}$ is $2$ . Since, the valency factor of $K M n O_{4}$ is $5$ in acidic medium, the number of moles required to reduce is $\frac{5}{2}$ moles.
Option B is correct.

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