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Calculate the mass of 95% pure CaCO3CaCO_3 that will be required to neutralize 50 mL of 0.5 M HCl solution according to the following reaction: CaCO3(s)+2HCl(aq)→CaCl2(aq)+CO2(g)+2H2O(l)CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + CO_2(g) + 2H_2O(l) [Calculate up to the second place of decimal point]

A

9.50 g

B

1.25 g

C

1.32 g

D

3.65 g

Step-by-Step Solution

  1. Calculate Moles of HCl: Using the molarity formula [Class 12 Chemistry, Ch 1, Eq. 1.8]: Moles of HCl=Molarity×Volume (L)\text{Moles of HCl} = \text{Molarity} \times \text{Volume (L)} nHCl=0.5 mol L−1×0.050 L=0.025 moln_{HCl} = 0.5 \text{ mol L}^{-1} \times 0.050 \text{ L} = 0.025 \text{ mol}
  2. Determine Stoichiometry: From the balanced equation, 1 mol CaCO31 \text{ mol } CaCO_3 reacts with 2 mol HCl2 \text{ mol } HCl. Moles of pure CaCO3=12×nHCl=0.0252=0.0125 mol\text{Moles of pure } CaCO_3 = \frac{1}{2} \times n_{HCl} = \frac{0.025}{2} = 0.0125 \text{ mol}
  3. Calculate Mass of Pure CaCO3CaCO_3: Molar mass of CaCO3=40+12+(3×16)=100 g mol−1CaCO_3 = 40 + 12 + (3 \times 16) = 100 \text{ g mol}^{-1}. Masspure=0.0125 mol×100 g mol−1=1.25 g\text{Mass}_{pure} = 0.0125 \text{ mol} \times 100 \text{ g mol}^{-1} = 1.25 \text{ g}
  4. Adjust for Purity: The sample is 95% pure, meaning 0.95×Masssample=Masspure0.95 \times \text{Mass}_{sample} = \text{Mass}_{pure}. Masssample=1.25 g0.95≈1.3157 g\text{Mass}_{sample} = \frac{1.25 \text{ g}}{0.95} \approx 1.3157 \text{ g} Rounding to two decimal places, the required mass is 1.32 g.
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