What volume of ${\mathrm{CO}}_2$ at STP will evolve when 1 g of ${\mathrm{CaCO}}_3$, reacts with excess of dil $\mathrm{HCl}$?

${\mathrm{CaCO}}_3+\mathrm{HCl}\to {\mathrm{CO}}_2+{\mathrm{CaCl}}_2+{\mathrm{H}}_2\mathrm{O}$

Above reaction states that one mole of CaCO₃ forms one mole of CO₂.

$$\begin{gathered} {\mathrm{n}}_{{\mathrm{CaCO}}_3}=1\times {\mathrm{n}}_{{\mathrm{CO}}_2} \\ {\left(\frac{\mathrm{mass}}{\mathrm{molar} \mathrm{mass}}\right)}_{{\mathrm{CaCO}}_3}={\left(\frac{\mathrm{mass}}{\mathrm{molar} \mathrm{mass}}\right)}_{{\mathrm{CO}}_2} \\ \frac{1 \mathrm{g}}{100 \mathrm{g}/\mathrm{mol}}=\frac{\mathrm{mass}}{44 \mathrm{g}/\mathrm{mol}} \\ \mathrm{mass}=0.44 \mathrm{g} \end{gathered}$$

Moles of CO₂

$\mathrm{moles}=\frac{0.44 \mathrm{g}}{44 \mathrm{g}/\mathrm{mol}}=0.01 \mathrm{mol}$

Volume of CO₂

$$\begin{gathered} 1 \mathrm{mol} \mathrm{of} {\mathrm{CO}}_2=22400 \mathrm{mL} \\ 0.01 \mathrm{mole} \mathrm{of} {\mathrm{CO}}_2=224 \mathrm{mL} \end{gathered}$$