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:

CaCO_3(s) + 2 HCl_(aq) ? CaCl_2(aq) + CO_2(g) + H₂O_(l)

The excess HCl(aq) is titrated by 7.65 mL of 0.125 M NaOH(aq). Calculate the mass percentage of CaCO3(s) in the sample.

Answer 1

7.65 mL of 0.125 M NaOH

(7.65 X 0.125) mmoles of NaOH = 0.956 mmoles of NaOH, which is the excess of HCl.

Now,

50.0 mL of 0.150 M HCl(aq)

(50 x 0.150) mmoles of HCl = 7.5 mmoles of HCl

Hence, HCl reacted with CaCO3 = 7.5 - 0.956 = 6.544 mmoles = 0.006544 moles

CaCO₃ + 2 HCl → CaCl₂ + CO₂ + H₂O

Supposing none of the impurities react with HCl:

(0.006544 mol HCl) x (1 mol CaCO3 / 2 mol HCl) x (100.0875 g CaCO3/mol) /(0.450 g)

= 0.7277

= 72.77%