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Q.

Decomposition of 50 grams of pure ${CaCO}_{3}$ liberates $x {dm}^{3}$ of ${CO}_{2}$ at STP. Mass of 80% pure KOH required for the complete neutralisation of the liberated gas (at STP) is

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a

42 g

b

70 g

c

84 g

d

56g

answer is D.

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Detailed Solution

$\underset{50 g}{{CaCO}_{3}} \to_{}^{} CaO + \underset{11.2 l}{{CO}_{2}}$
$\underset{2 (56 g)}{2 KOH} + \underset{\begin{array}{l} 22.4 l \\ 11.2 l \end{array}}{{CO}_{2}} \to_{}^{} K_{2} {CO}_{3} + H_{2} O$
$W_{KOH} = 56 g$
$W_{KOH} (80 % pure) = 56 \times \frac{100}{80} = 70 g$

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