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Q.

In the following reaction, ${MnO}_{2} + 4 HCl ⟶ {MnCl}_{2} + 2 H_{2} O + {Cl}_{2}$
$2 moles of {MnO}_{2} react with 4 moles of HCl to form$
$11.2 {LCl}_{2} at STP.$
$Thus, per cent yield of {Cl}_{2} is$

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a

75%

b

25%

c

50%

d

100%

answer is B.

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Detailed Solution

$\underset{2 mol}{{MnO}_{2}} + 4 \underset{4 mol}{4 HCl} ⟶ {MnCl}_{2} + 2 H_{2} O + \underset{1}{{Cl}_{2}}$

But the yield is 11.2

$∴ Percentage yield = \frac{11.2}{22.4} \times 100 = 50 %$

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