Q. $0.32g$ of $CaC_{2}$ reacts with water produces gas $X$ . The gas $X$ combination with $O_{2}$ produce $CO_{2}$ . The volume of $CO_{2}$ at STP:

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A

$112mL$

B

$484mL$

C

$224mL$

D

$448mL$

Solution:

$\left(CaC\right)_{2}+2H_{2}O \rightarrow Ca\left(\right.OH\left(\left.\right)_{2}+C_{2}H_{2}$
$\frac{0 . 32}{64}=0.005mole=5mmoleCaC_{2}$ or $C_{2}H_{2}$
$=10mmoleCO_{2}$
$=224ml$ of $CO_{2}$

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