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

Step 1: Understanding the Concept:
This is a stoichiometry problem involving a neutralization reaction. We must find the amount of pure reactant needed and then adjust for its percentage purity in the actual sample.
Key Formula or Approach:
1. Moles \(= \text{Molarity} \times \text{Volume (in L)}\).
2. Mole ratio from balanced equation: \(1 \text{ mole } CaCO_3 : 2 \text{ moles } HCl\).
3. \(\text{Mass} = \text{Moles} \times \text{Molar Mass}\).
4. \(\text{Required Sample Mass} = \frac{\text{Pure Mass}}{\text{Purity Percentage}} \times 100\).
Step 2: Detailed Explanation:
1. Calculate moles of HCl:
\[ n_{HCl} = 0.5 \text{ mol/L} \times \frac{50}{1000} \text{ L} = 0.025 \text{ moles} \]
2. Find moles of pure \(CaCO_3\) required:
From the reaction: \(1 \text{ mol } CaCO_3 \equiv 2 \text{ mol } HCl\).
\[ n_{CaCO_3} = \frac{1}{2} \times n_{HCl} = \frac{0.025}{2} = 0.0125 \text{ moles} \]
3. Calculate mass of pure \(CaCO_3\):
Molar mass of \(CaCO_3 = 40 + 12 + (3 \times 16) = 100 \text{ g/mol}\).
\[ \text{Mass}_{\text{pure}} = 0.0125 \times 100 = 1.25 \text{ g} \]
4. Adjust for 95% purity:
Since only 95% of the sample is \(CaCO_3\):
\[ \text{Mass}_{\text{sample}} \times \frac{95}{100} = 1.25 \text{ g} \]
\[ \text{Mass}_{\text{sample}} = \frac{1.25 \times 100}{95} = 1.3157... \text{ g} \]
Rounding to two decimal places, we get 1.32 g.
Step 3: Final Answer:
The required mass of 95% pure \(CaCO_3\) is 1.32 g.