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 One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.