For decolorisation of $1 m o l e$ of $K M n O_{4}$ , the moles of $H_{2} O_{2}$ required is

\frac{1}{2}

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\frac{3}{2}

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\frac{5}{2}

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\frac{7}{2}

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Solution

The correct option is A $\frac{1}{2}$

Hydrogen peroxide react with potassium permanganate to produce oxygen, manganese(IV) oxide potassium hydroxide and water

$3 H_{2} O_{2} + 2 K M n O_{4} \to 3 O_{2} + 2 M n O_{2} + 2 K O H + 2 H_{2} O$
According to the equation 2 moles of $H_{2} O_{2}$ reacts with 3 moles of $K M n O_{4}$
So 3/2 moles of $H_{2} O_{2}$ will react with 1 moles of $K M n O_{4}$ .
Hence option A is correct.

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