Question

A 4.51 g sample of chalk is mixed with 23.8 mL of 3.0 M hydrochloric acid. It is filtered to obtain 3.64 g of calcium carbonate. What is the percent of calcium carbonate in the chalk?

Answer 1

No.of moles of HCl reacted = (23.8 x 10^-3 L) x (3.0 M) = 0.0714 mol

Calcium carbonate reacts with HCl as shown below:

CaCO₃ + 2 HCl ------------> CaCl₂ + CO₂ + H₂O

From this balanced reaction, we have:

1 mole of CaCO₃ reacts with 2 moles of HCl

Hence, 1 mole of HCl reacts with 1/2 mole of CaCO₃

Therefore, 0.0714 mol of HCl reacts with (1/2) x 0.0714 mol mole of CaCO₃ = 0.0357 mol mole of CaCO₃

Molar mass of CaCO₃= 100.0869 g/mol

Hence, Mass of CaCO₃ in chalk = (0.0357 mol) x ( 100.0869 g/mol) = 3.573 g

Mass of chalk = 4.51 g

Hence, percent of calcium carbonate in the chalk will be:

percentage = (3.573 g) / (4.51 g) x 100 %

percentage = 0.7923 x 100 %

percentage = 79.23 %

• Percent of calcium carbonate in the chalk = 79.23 %