Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction, CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)

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

≡ [(0.75 mol) × (36.5 g mol–1)] HCl is present in 1 L of water

≡ 27.375 g of HCl is present in 1 L of water

Thus, 1000 mL of solution contains 27.375 g of HCl.

Amount of HCl present in 25 mL of solution, by calculation

= 0.6844 g

From the given equation,

2 mol of HCl (2 × 36.5 = 71 g) react with 1 mol of CaCO3 (100 g).

Amount of CaCO3 that will react with 0.6844 g

= 0.9639 g