Moles of HCl used \(= 0.1\times0.040=0.004\,\text{mol}\). Reaction of HCl with \(\text{Ca(OH)}_2\): \(\text{Ca(OH)}_2+2\text{HCl}\rightarrow\text{CaCl}_2+2\text{H}_2\text{O}\). So moles of unreacted \(\text{Ca(OH)}_2=\frac{0.004}{2}=0.002\,\text{mol}\). Since \(\text{CO}_2\) reacted with exactly half the initial \(\text{Ca(OH)}_2\), the reacted half also \(=0.002\,\text{mol}\), and initial \(\text{Ca(OH)}_2=0.004\,\text{mol}\). Reaction: \(\text{CO}_2+\text{Ca(OH)}_2\rightarrow\text{CaCO}_3+\text{H}_2\text{O}\) (1:1 molar ratio). So moles of \(\text{CO}_2\) reacted \(=0.002\,\text{mol}\). At STP (1 atm, 273 K): volume of 1 mol ideal gas \(=22400\,\text{cm}^3\). Volume \(x=0.002\times22400=44.8\approx45\,\text{cm}^3\).