44 g CO2.........6,022 * 10 la 23 atomi
88 g CO2..............x atomi x = 12.044 * 10 la 23 atomi
1 mol MgCl2.........2*6,022 * 10 la 23 ioni Cl -
2 moli MgCl2..................x ioni Cl - x = 24,088 * 10 la 23 ioni Cl-
nr moli = masa/masa molara => masa CaCO3 = 4 * 100 = 400 g