$\mathrm{CaCO}_{3(\mathrm{s})} + 2\mathrm{HCl}_{(\mathrm{aq})}$ → $\mathrm{CaCl}_{2(\mathrm{aq})} + \mathrm{CO}_{2(\mathrm{g})} + \mathrm{H}_2\mathrm{O}_{(\mathrm{l})}$ No. of moles of HCl given = $M_{\mathrm{HCl}} V_{\mathrm{HCl}}$ = 0.75 mole $L^{-1}$ × 25 × $10^{-3}$ L = 18.75 × $10^{-3}$ mole = 0.0188 mole 2 moles of HCl requires 1 mole of $\mathrm{CaCO}_3$ 0.0188 mole of HCl will require = $\frac{0.0188}{2}$ = 0.0094 mole of $\mathrm{CaCO}_3$ Molar mass of $\mathrm{CaCO}_3$ = 100 g/mole Mass of $\mathrm{CaCO}_3$ required = 100 × 0.0094 = 0.94 g