Reaction between $\mathrm{CO_2}$ and lime water

The reaction is: $\mathrm{Ca(OH)_2 + CO_2 \rightarrow CaCO_3 + H_2O}$

Thus, $1$ mole of $\mathrm{CO_2}$ reacts with $1$ mole of $\mathrm{Ca(OH)_2}$.

Given condition

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

The question says that the entire amount of $\mathrm{CO_2}$ reacted with exactly half the initial amount of $\mathrm{Ca(OH)_2}$.

Therefore, moles of $\mathrm{CO_2}$ passed: $n_{\mathrm{CO_2}}=\frac{n}{2}$

So the unreacted $\mathrm{Ca(OH)_2}$ left after absorption is also: $n-\frac{n}{2}=\frac{n}{2}$

Neutralization of remaining $\mathrm{Ca(OH)_2}$ with HCl

Neutralization reaction: $\mathrm{Ca(OH)_2 + 2HCl \rightarrow CaCl_2 + 2H_2O}$

Given HCl used = $0.1\,\mathrm{M}$, $40\,\mathrm{mL}$

Moles of HCl: $n_{\mathrm{HCl}}=0.1\times 0.040=0.004$

From stoichiometry, $n_{\mathrm{Ca(OH)_2\,left}}=\frac{0.004}{2}=0.002$

But this remaining amount is $\dfrac{n}{2}$.

Hence, $\frac{n}{2}=0.002 \Rightarrow n=0.004$

Moles of $\mathrm{CO_2}$

Since $\mathrm{CO_2}$ reacted with half the initial $\mathrm{Ca(OH)_2}$: $n_{\mathrm{CO_2}}=\frac{n}{2}=0.002\,\text{mol}$

Use ideal gas equation at $273\,\mathrm{K}$ and $1\,\mathrm{atm}$

At STP, $1$ mole of gas occupies $22.4\,\mathrm{L}$.

Therefore volume of $0.002$ mol $\mathrm{CO_2}$: $V=0.002\times 22.4=0.0448\,\mathrm{L}$

Convert to $\mathrm{cm^3}$: $0.0448\,\mathrm{L}=44.8\,\mathrm{cm^3}$

Nearest integer: $x=45\,\mathrm{cm^3}$

Comparison with stored answer

Stored correct answer = $45$

Our derived answer also equals $45$.