Calcium carbonate reacts with aqueous HCl to give

Solution:

0.75 M of HCl ≡ 0.75 mol of HCl are present in 1 L of water

$\equiv\left[(0.75 \mathrm{~mol}) \times\left(36.5 \mathrm{~g} \mathrm{~mol}^{-1}\right)\right] \mathrm{HCl}$ is present in $1 \mathrm{~L}$ of water

$\equiv 27.375 \mathrm{~g}$ of $\mathrm{HCl}$ is present in $1 \mathrm{~L}$ of water

Thus, $1000 \mathrm{~mL}$ of solution contains $27.375 \mathrm{~g}$ of $\mathrm{HCl}$.

$\therefore$ Amount of $\mathrm{HCl}$ present in $25 \mathrm{~mL}$ of solution

$=\frac{27.375 \mathrm{~g}}{1000 \mathrm{~mL}} \times 25 \mathrm{~mL}$

$=0.6844 \mathrm{~g}$

From the given chemical equation,

$\mathrm{CaCO}_{3(s)}+2 \mathrm{HCl}_{(a q)} \longrightarrow \mathrm{CaCl}_{2(\alpha))}+\mathrm{CO}_{2(g)}+\mathrm{H}_{2} \mathrm{O}_{(i)}$

$2 \mathrm{~mol}$ of $\mathrm{HCl}(2 \times 36.5=71 \mathrm{~g})$ react with $1 \mathrm{~mol}$ of $\mathrm{CaCO}_{3}(100 \mathrm{~g})$.

$\therefore$ Amount of $\mathrm{CaCO}_{3}$ that will react with $0.6844 \mathrm{~g}=\frac{100}{71} \times 0.6844 \mathrm{~g}$

$=0.9639 \mathrm{~g}$