Chemistry - Mole Concept Question with Solution | TestHub

Let the initial moles of $\mathrm{Ca}(\mathrm{OH}{)}_2$ be $n_0$.

${\mathrm{CO}}_2$ reacts with half of $\mathrm{Ca}(\mathrm{OH}{)}_2$:

${\mathrm{CO}}_2+\mathrm{Ca}(\mathrm{OH}{)}_2\to {\mathrm{CaCO}}_3+{\mathrm{H}}_2\mathrm{O}$

Moles of ${\mathrm{CO}}_2$ = Moles of $\mathrm{Ca}(\mathrm{OH}{)}_2$ reacted = $n_0/2$.

Excess $\mathrm{Ca}(\mathrm{OH}{)}_2$ = $n_0-n_0/2=n_0/2$.

This excess is neutralized by HCl:

$\mathrm{Ca}(\mathrm{OH}{)}_2+2\mathrm{HCl}\to {\mathrm{CaCl}}_2+2{\mathrm{H}}_2\mathrm{O}$

Moles of HCl = $0.1M\times 0.040L=0.004mol$.

Moles of $\mathrm{Ca}(\mathrm{OH}{)}_2$ reacted with HCl = $0.004/2=0.002mol$.

So, $n_0/2=0.002mol$.

Moles of ${\mathrm{CO}}_2$ = $0.002mol$.

Using the ideal gas law, $PV=nRT$:

$1atm\times V=0.002mol\times 0.0821Latmmo{l}^{-1}{K}^{-1}\times 273K$

$V=0.0448L=44.8c{m}^3$.

Nearest integer is $45c{m}^3$.

The final answer is $\boxed{45}$.