If 1 gm of HCl and 1 gm of ${MnO}_{2}$ heated together, the maximum weight of ${Cl}_{2}$ gas evolved will be :
$[{MnO}_{2} + 4 HCl ⟶ {MnCl}_{2} + {Cl}_{2} + 2 H_{2} O]$
[Atomic mass of Mn = 55]

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2 gm

0.975 gm

0.486 gm

0.972 gm

answer is C.

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Detailed Solution

In the given reaction

$\begin{array}{l} {MnO}_{2} + 4 HCl \to \underline{{MnCl}_{2}} + \underline{{Cl}_{2}} + \underline{2 H_{2} O} \\ n = \frac{1}{87} \frac{1}{36.5} \\ (L.R.) \frac{1}{4} \times \frac{1}{36.5} \times 71 = 0.486 g \end{array}$

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