Question

A 0.450 gram sample of impure CaCO3 (s) is dissolved in 50.0 mL of 0.150 M HCl (aq). The equation for the reaction is CaCO3(s) + 2HCl --> CaCL2 (aq) + H2O (l) + CO2 (g). The excess HCl (aq) is titrated by 7.20 mL of 0.125 M NaOH (aq). Calculate the mass percentage of CaCO3 (s) in the sample.

Answer 1

Moles / Liter = Molarity ------> mol = Molarity * Volume in liters

Excess HCl ---> x / 0.00720 L = 0.125 M                                                 (7.20 mL = 0.00720 L)
x = 0.0009 mol

Total HCl ----> x / 0.05 L = 0.150 M                                                          (50.0 mL = 0.05 L)
x = 0.0075 mol

Actual used HCl = Total HCl - Excess HCl mol = 0.0075 - 0.0009 mol = 0.0066 mol
mass of CaCO₃ = 0.0066 mol HCl X 1mol CaCO₃ / 2mol HCl X 100.0869 g/1 mol CaCO3 = 0.3303 g

The mass of CaCO₃ (s) in the sample = 0.3303 g
mass / Total mass = (0.3303 g / 0.450 g) * 100 = 73.4%

mass percentage of CaCO3 = 73.4 %