Question

What mass of 95% pure \(CaCO_3\) will be required to neutralise 50 mL of 0.5 M \(HCl\) solution according to the following reaction?
\(CaCO_3(s) + 2HCl(aq) → CaCl_2(aq) + CO_2(g) + 2H_2O(l)\)
[Calculate upto second place of decimal point]

• A. 1.25 g
• B. 1.32 g
• C. 3.65 g
• D. 9.50 g

Answer 1

The Correct Option is B

To determine the mass of 95% pure \(CaCO_3\) required to neutralise 50 mL of 0.5 M \(HCl\), we need to follow these steps:

1. Understand the reaction: The chemical equation given is \(CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + 2H_2O\). From this, we see that 1 mole of \(CaCO_3\) reacts with 2 moles of \(HCl\).

2. Calculate moles of \(HCl\): Using the formula \(Moles = Molarity \times Volume\ (in\ L)\), we have:

\[0.5 \, \text{M} \times 0.050 \, \text{L} = 0.025 \, \text{moles of } HCl\]

3. Determine moles of \(CaCO_3\) required: From the stoichiometry of the reaction, \(2 \, \text{moles of} \, HCl\) require \(1 \, \text{mole of} \, CaCO_3\). Therefore:

\[0.025 \, \text{moles } HCl \times \frac{1 \, \text{mole } CaCO_3}{2 \, \text{moles } HCl} = 0.0125 \, \text{moles of } CaCO_3\]

4. Calculate mass of pure \(CaCO_3\): Using the molar mass of \(CaCO_3\) (100 g/mol), the mass is:

\[0.0125 \, \text{moles } \times 100 \, \text{g/mol} = 1.25 \, \text{g}\]

5. Adjust for purity: Since the sample is 95% pure, the actual mass required is:

\[1.25 \, \text{g} \div 0.95 = 1.32 \, \text{g}\]

The mass of 95% pure \(CaCO_3\) required is 1.32 g.