Question: 51: 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?
\mathrm{CaCO}_{3(\mathrm{~s})}+2 \mathrm{HCl}_{(\mathrm{aq})} \rightarrow \mathrm{CaCl}_{2(\mathrm{aq})}+\mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{i})}[Calculate upto second place of decimal point]
(1) 3.65 g
(2) 9.50 g
(3) 1.25 g
(4) 1.32 g
Answer: Option (4)
Explanation: Moles of HCl present in the solution are calculated using molarity.
\text{Moles of HCl} = 0.5 \times \frac{50}{1000} \text{Moles of HCl} = 0.025From the balanced chemical equation, 2 moles of HCl react with 1 mole of CaCO3.
Therefore, moles of CaCO3 required are:
\text{Moles of CaCO}_{3} = \frac{0.025}{2} \text{Moles of CaCO}_{3} = 0.0125Molar mass of CaCO3 is 100 \, \text{g mol}^{-1}.
Mass of pure CaCO3 required is:
\text{Mass} = 0.0125 \times 100 \text{Mass} = 1.25 \, \text{g}Since the sample is only 95 \% pure, the actual mass required will be higher.
\text{Required mass} = \frac{1.25 \times 100}{95} \text{Required mass} = 1.32 \, \text{g}Thus, the correct answer is option (4).