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The decomposition of a certain mass of ${CaCO}_{3}$ gave 11 .2 ${dm}^{3}$ of ${CO}_{2}$ gas at STP. The mass of KOH required to completely neutralise the gas is

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56 g

28 g

42 g

20 g

answer is B.

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Detailed Solution

$KOH + {CO}_{2} \to {KHCO}_{3}$

39+16+1 22.4 ${dm}^{3}$

= 56 g

$∵ For 22.4 {dm}^{3} {CO}_{2} required KOH = 56 g$

$∴ For 11.2 {dm}^{3} {CO}_{2} will require KOH = \frac{56 \times 11.2}{22.4} = 28 g$

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