The volume of carbon dioxide gas evolved at S.T.P. by heating $7.3 g m$ of $M g {({H C O}_{3})}_{2}$ will be:

2240 m l

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1120 m l

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2340 m l

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2000 m l

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Solution

The correct option is A $2240 m l$
$M g (H C O_{3})_{2} ⟶ M g O + H_{2} O + 2 C O_{2}$
Molecular mass of $M g (H C O_{3})_{2} = 146$ & that of $C O_{2} = 44$
$∴$ Number of moles of $M g (H C O_{3})_{2} = \frac{7.3}{146} = 0.05$
$∴$ Number of moles of $C O_{2} = 2 \times 0.05 = 0.1$
At STP, $1$ mole of any gas has volume of $22.4 L$
$∴$ $0.1$ mole of $C O_{2}$ occupies= $22.4 \times 0.1$
$\Rightarrow 2.24 L = 2240 m L$

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