Stoichiometry question?

Question

To test for chalk, a soil sample (4.5g) was dissolved in 9.0mL of HCl (1.00mol/L). The unused acid was titrated against 1.0 molar NaOH(aq); titre value 3.50mL. Calculate the chalk (CaCO3) content as % by mass in the soil sample.

Answer 1

5.5 mL of HCl was consumed.

CaCO3 + 2HCl -> CaCl2 + H2O + CO2

so, moles of CaCO3 = 5.5 x 10^-3 / 2 moles = 2.75 mmol

Thus %age of CaCO3 = (2.75x10^-3x100 / 4.5 ) x 100 = 6.11 %

Answer 2

6.11% of caco3