$2 L i O H (s) + {C O}_{2} (g) \to {L i}_{2} {C O}_{3} (s) + H_{2} O (l)$
If $0.96 g$ of $L i O H$ is to react fully with ${C O}_{2}$ , what volume of ${C O}_{2}$ would be required? Assume STP conditions.

450 m L

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350 m L

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500 m L

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300 m L

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Solution

The correct option is A $450 m L$
Given, $2 L i O H + C O_{2} \to L i_{2} C O_{3} + H_{2} O$

At STP to react with $2 m o l e s$ of $L i O H$ , we need $22.4 L i t r e s$ of $C O_{2}$

$2 m o l e s$ of $L i O H$ $= 2 \times (7 + 16 + 1) = 48 g$ $L i O H$

$48 g L i O H$ $\to 22.4 L C O_{2}$
$0.96 g \to x$
$\Rightarrow x = \frac{22.4 L \times 0.96}{48} = 448 m L$
~450 ml is the correct answer

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