44.8 L of $C O_{2}$ at NTP is obtained by heating $x$ g of pure $C a C O_{3} . x$ is:

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Solution

$C a C O_{3} ⟶ C a O + C O_{2}$
no. of moles of $C O_{2}$ in $44.8 L$ at NTP.
No. of moles $= \frac{Volume of gas}{Volume at NTP}$
$n = \frac{44.8 L}{22.4 L} = 2$ moles of $C O_{2}$ .
As $1$ mole of $C O_{2}$ is produced by $1$ mole of $C a C O_{3}$ thus,
$2$ moles of $C O_{2}$ is produced by $2$ moles of $C a C O_{3}$ .
Amount of $C a C O_{3}$ present in $2$ mole of $C a C O_{3}$ ,
$m = 2 m o l e \times 100 g m o l^{- 1}$
$n = 200 g$

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