When one mole of $K M n O_{4}$ reacts with $H C l$ , the volume of chlorine liberated at NTP will be :

11.2 L

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22.4 L

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44.8 L

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56.0 L

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Solution

The correct option is C $56.0 L$
$2 K M n O_{4} + 16 H C l \to 2 K C l + 2 M n C l_{2} + 8 H_{2} O + 5 C l_{2}$
So, one mole of $K M n O_{4}$ will give $22.4 \times 2.5 = 56 L C l_{2}$

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