$0.800 g$ of impure $CaCO X_{3}$ is heated with $50.0 mL$ of $0.200 M$ $HCl$ (excess). The unreacted acid needs $12.0 mL$ of $0.250 M$ $NaOH$ . Taking $CaCO X_{3} = 100$ , the percentage of $CaCO X_{3}$ in the sample is:

Answer: C

Total HCl

= 50.0 \times 0.200 = 10.0

mmol; excess

= 12.0 \times 0.250 = 3.0

mmol; HCl used

= 7.0

mmol. Since

CaCO X_{3} + 2 HCl

CaCO X_{3}

= 3.5

mmol

= 350

mg. Percentage

= 350/800 \times 100 = 43.75%