What mass of \( \text{CaCO}_3 \) is required to react completely with \( 25\text{ mL} \) of \( 0.75\text{ M} \text{ HCl} \)? (Atomic masses: \( \text{Ca} = 40, \text{C} = 12, \text{O} = 16 \))

Reaction: \( \text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2 \). Moles of HCl = \( 0.025 \times 0.75 = 0.01875\text{ mol} \). Moles of \( \text{CaCO}_3 = 0.01875 / 2 = 0.009375\text{ mol} \). Mass = \( 0.009375 \times 100 = 0.9375\text{ g} \dots \) wait, \( 0.938 / 2 = 0.469\text{ g} \).