A $0.900 g$ sample of limestone (only $CaCO X_{3}$ reacts with acid) is treated with $40.0 mL$ of $0.500 N$ $HCl$ . The excess acid requires $24.0 mL$ of $0.250 N$ $NaOH$ for back titration. Find the percentage of $CaCO X_{3}$ in the sample (one decimal; $CaCO X_{3} = 100$ ).

Numerical answer type, answer in %, tolerance 0.1.

Answer: 77.8 %

Total acid

= 40.0 \times 0.500 = 20.0

meq; excess acid

= 24.0 \times 0.250 = 6.0

meq. Acid used by

CaCO X_{3} = 14.0

meq. Equivalent mass of

CaCO X_{3} = 100/2 = 50

, so mass

= 14.0 \times 50 = 700

mg. Purity

= 700/900 \times 100 = 77.8%