Question

The mass of \(CaCO_3\) required to react with 25 mL of 0.75 M HCl is

• A. 0.94 g
• B. 9.4 g
• C. 0.094 g
• D. 0.49 g

Hint:

Always write the balanced chemical equation first to get the correct mole ratio.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The reaction between calcium carbonate and hydrochloric acid is: \(CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2\).
Step 2: Detailed Explanation:
Moles of HCl = Molarity × Volume (in L) = \(0.75 \times 0.025 = 0.01875\) mol. From the balanced equation, 2 moles of HCl react with 1 mole of \(CaCO_3\). Moles of \(CaCO_3\) required = \(\frac{0.01875}{2} = 0.009375\) mol. Molar mass of \(CaCO_3\) = \(40 + 12 + (3 \times 16) = 100\) g/mol. Mass of \(CaCO_3\) = \(0.009375 \times 100 = 0.9375\) g $\approx$ 0.94 g.
Step 3: Final Answer:
The mass required is 0.94 g, option (A).