To 20 g of CaCO3, 400 cm3 of HCl solution with a molar concentration of 6 mol / dm3 was added.

To 20 g of CaCO3, 400 cm3 of HCl solution with a molar concentration of 6 mol / dm3 was added. Calculate the chemical amount of H + ions in the resulting solution (complete dissociation).

СaCO3 + 2HCl = CaCl2 + H2O + CO2
V (HCl solution) = 400 cm3 = 0.4 l;
n (HCl) = V (HCl solution) * C (HCl) = 0.4 * 6 = 2.4 mol;
n (CaCO3) = m (CaCO3) / M (CaCO3) = 20/100 = 0.2 mol;
Based on the stoichiometry of the reaction, 0.4 mol of HCl will react with calcium carbonate. The rest of the acid will remain in solution. neglecting the dissociation of water, we get:
n (H +) = n (unreacted HCl) = 2.4 – 0.4 = 2 mol.
Answer: n (H +) = 2 mol.