The number of moles of ${\mathrm{CO}}_2$ produced when 3 moles of $\mathrm{HCl}$ is treated with excess of ${\mathrm{CaCO}}_3$ is

        ${\mathrm{CaCO}}_3+2\mathrm{HCl}\to {\mathrm{CaCl}}_2+{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2$

Here, 2 moles of $\mathrm{HCl}$ reacts with 1 mole ${\mathrm{CaCO}}_3$ to give 1 mole of ${\mathrm{CO}}_2$.

So, 1 mole $\mathrm{HCl}=1/2$ mole of ${\mathrm{CO}}_2$

3 moles of $\mathrm{HCl}=3\times \frac{1}{2}=1.5$ moles of ${\mathrm{CO}}_2$

Therefore, 3 moles of $\mathrm{HCl}$ gives $1.5$ moles of ${\mathrm{CO}}_2$.