How many grams of calcium carbonate (CaCO_3) are required to completely react with 25.0 mL of 0.750 M hydrochloric acid (HCl)?

• A. 0.94 g
• B. 1.50 g
• C. 0.75 g
• D. 2.10 g

Answer: (A)

Verified step by step guidance

1. Write the balanced chemical equation for the reaction between calcium carbonate and hydrochloric acid: \(\mathrm{CaCO_3 (s) + 2HCl (aq) \rightarrow CaCl_2 (aq) + CO_2 (g) + H_2O (l)}\).
2. Calculate the moles of hydrochloric acid (HCl) using the formula: \(\text{moles HCl} = M \times V\), where \(M = 0.750\, \mathrm{mol/L}\) and \(V = 25.0\, \mathrm{mL} = 0.0250\, \mathrm{L}\).
3. Use the stoichiometric ratio from the balanced equation to find the moles of calcium carbonate (CaCO_3) needed. Since 2 moles of HCl react with 1 mole of CaCO_3, calculate: \(\text{moles CaCO_3} = \frac{\text{moles HCl}}{2}\).
4. Calculate the mass of calcium carbonate required by multiplying the moles of CaCO_3 by its molar mass. The molar mass of CaCO_3 is calculated as: \$40.08 + 12.01 + (3 \times 16.00) = 100.09\, \mathrm{g/mol}\(, so \)\text{mass CaCO_3} = \text{moles CaCO_3} \times 100.09\, \mathrm{g/mol}$.
5. The resulting mass from the calculation will give the grams of calcium carbonate required to completely react with the given hydrochloric acid solution.