What mass of 95% pure CaCO3 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) [Calculate upto second place of decimal point]
- 1.25 g
- 1.32 g
- 3.65 g
- 9.50 g
Answer
(2) 1.32 g
Solution
Moles of HCl = 0.5 501000 = 0.025 mol
From the equation, 2 mol HCl ≡ 1 mol CaCO3
⇒ moles of CaCO3 = 0.0252 = 0.0125 mol
Mass of pure CaCO3 = 0.0125 100 = 1.25 g
∵ the sample is 95% pure,
mass of sample = 1.25 10095
= 1.3158 g 1.32 g
Answer: (2)