How much CaCO3 is required to obtain 571.2 liters of CO2 by reaction with hydrochloric acid?

Let us carry out the solution, according to the condition of the problem, we write the equation:
X g .-? V = 571.2 liters.
1. CaCO3 + 2HCl = CaCl2 + H2O + CO2 – ion exchange, carbon monoxide is released (4);
2. Let’s make calculations using the formulas:
M (CaCO3) = 100 g / mol;
M (CO2) = 44 g / mol.
3. Let’s make the proportion:
1 mol of gas at normal level – 22.4 liters;
X mol (CO2) – 571.2 liters. hence, X mol (CO2) = 1 * 571.2 / 22.4 = 25.5 mol;
Y (CaCO3) = 25.5 mol since the amount of these substances is 1 mol.
4. Find the mass of the original substance:
m (CaCO3) = Y * M = 25.5 * 100 = 2550 g = 2 kg. 550 BC
Answer: to carry out the process, a CaCO3 salt weighing 2 kg is required. 550 BC