How much hydrochloric acid and calcium carbonate is required to produce 2.8 liters of carbon dioxide?

To solve the problem, we write the equation:
2HCl + CaCO3 = CaCl2 + CO2 + H2O – ion exchange reaction, carbon monoxide is released (4);
Let’s calculate the molar masses:
M (HCl) = 36.5 g / mol;
M (CaCO3) = 100 g / mol;
M (CO2) = 44 g / mol;
Let’s calculate the amount of mol of CO2, HCl:
1 mol of gas at normal level – 22.4 liters.
X mol (CO2) – 2.8 liters. hence, X mol (CO2) = 1 * 2.8 / 22.4 = 0.125 mol;
X mol (HCl) – 0.125 mol (CO2);
-2 mol – 1 mol from here, X mol (HCl) = 2 * 0.125 / 1 = 0.25 mol;
According to the equation, the number of moles CaCO3, CO2 is equal to 1 mol, which means that Y (CaCO3) = 0.125 mol.
We find the mass of salt, hydrochloric acid:
m (CaCO3) = Y * M = 0.125 * 100 = 12.5 g;
m (HCl) = Y * M = 0.25 * 36.5 = 9.125 g.
Answer: the mass of calcium carbonate is 12.5 g, the mass of hydrochloric acid is 9.125 g.