When $o n e g r a m m o l e$ of $K M n O_{4}$ is mixed with hydrochloric acid then, the volume of chlorine gas liberated at $N T P$ will be:

67.2 l i t r e

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22.4 l i t r e

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56 l i t r e

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44.8 l i t r e

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Solution

The correct option is A $67.2 l i t r e$
$\begin{matrix} {K M n O_{4}}_{1 m o l e} + 8 H C l \to K C l + M n C l_{2} + 3 C l_{2} ↑ + 2 H_{2} O (b a l a n c e d e q u a t i o n) sin c e, 1 m o l e o f K M n O_{4} g i v e s 3 m o l e s o f C l_{2} g a s S o, 1 g = 22.4 L ∴ v o l u m e o f t h e C l_{2} g a s o b t i n e d = 3 \times 22.4 = 67.2 L H e n c e, t h e o p t i o n A i s t h e c o r r e c t a n s w e r . \end{matrix}$

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