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What mass of 95% pure CaCO3CaCO_3 will be required to neutralise 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 upto second place of decimal point]

1

1.32 g

2

1.38 g

3

1.45 g

4

1.52 g

Step-by-Step Solution

Moles of HCl = M×V(L)=0.5×0.050=0.025M \times V(L) = 0.5 \times 0.050 = 0.025 mol. From the stoichiometry, 1 mole of CaCO3CaCO_3 reacts with 2 moles of HCl. So, moles of CaCO3CaCO_3 required = 0.025/2=0.01250.025 / 2 = 0.0125 mol. Molar mass of CaCO3=40+12+3×16=100CaCO_3 = 40 + 12 + 3 \times 16 = 100 g/mol. Mass of pure CaCO3=0.0125×100=1.25CaCO_3 = 0.0125 \times 100 = 1.25 g. Since the sample is 95% pure, mass of sample required = 1.25/0.95≈1.31571.25 / 0.95 \approx 1.3157 g, which rounds to 1.32 g. Note: Based on the provided answer key (2), the calculation might imply a different rounding or specific molar mass precision, but 1.32 g is the standard result.

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