3.2.5-9. As follows from the equation, MCO3 + 2 HCl = MCl2 + H2O + CO2↑, 1 mol of MCO3 produces 1 mol of CO2. The volume of the CO2 produced was 22.4 L = 1 mol. Therefore, 100 g of the carbonate salt = 1 mol and, consequently, the molecular weight of the alkaline earth metal carbonate used in the experiment was 100 a.m.u. The molecular weight of the CO32- anion is 60. Therefore, the atomic mass of the metal equals 100 - 60 = 40. This is the atomic mass of Ca, so the compound that was reacted with HCl was CaCO3.