1.50 g of impure $CaCO X_{3}$ (molar mass 100 g/mol) is treated with 50.0 mL of 0.500 M HCl. The excess acid requires 18.0 mL of 0.250 M NaOH in a back titration. Find the percentage of $CaCO X_{3}$ in the sample (one decimal place; impurities do not react).

Numerical answer type, answer in %, tolerance 0.06.

Answer: 68.3 %

Total HCl = 25.0 mmol; excess HCl = 0.250

\times

18.0 = 4.50 mmol, so HCl used = 20.5 mmol. Each

CaCO X_{3}

consumes 2 HCl, giving 10.25 mmol = 1.025 g. Percentage = 1.025/1.50

\times

100 = 68.3 %.