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

March 24, 2021 | education

| 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

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