The mass of ${\mathrm{CaCO}}_3\left(\mathrm{s}\right)$ that reacts completely with $50\mathrm{mL}$ of  $0.75\mathrm{M} \mathrm{HCl}$ is  

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

The amount of $\mathrm{HCl}$ in $50\mathrm{mL}(=0.050\mathrm{L})$of $0.75 \mathrm{M}$ solution is

$\mathrm{n}=\mathrm{MV}=(0.75\mathrm{molL}=(0.050\mathrm{L})=0.0375\mathrm{mol}$ 

From the chemical equation, we find that the amount of ${\mathrm{CaCO}}_3$ will be $0.0375\mathrm{mol}/2=0.01875\mathrm{mol}$

Mass ${\mathrm{CaCO}}_3$ required will be $\mathrm{m}=\mathrm{nM}=(0.01875\mathrm{mol})\left(100\mathrm{g}{\mathrm{mol}}^{-1}\right)=1.875\mathrm{g}.$