Chlorine gas is prepared by reaction of $H_{2} S O_{4}$ with $M n O_{2}$ and NaCl. What volume of $C l_{2}$ will be produced at STP if 50 g of NaCl is taken in the reaction?

1.915 L

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22.4 L

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11.2 L

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9.57 L

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Solution

The correct option is D 9.57 L
Chemical reaction involved is:
$2 N a C l + 3 H_{2} S O_{4} + M n O_{2} \to 2 N a H S O_{4} + M n S O_{4} + C l_{2} + 2 H_{2} O$
Molar mass of $N a C l = 23 + 35.5 = 58.5 g$
According to the balanced reaction $58.5 \times 2 = 117 g$ of $N a C l$ produces 1 mol of $C l_{2}$ i.e. $22.4 L$ at STP.
$∴ 50 g N a C l$ will produce $\frac{22.4}{117} \times 50 = 9.57 L C l_{2}$ at STP
option D is correct

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