How many grams of CaCO3 are formed when 200 grams of Ca (OH) 2 react with 33.6 liters of CO2?

Let’s implement the solution:

According to the condition of the problem, we write down the equation of the process:
Ca (OH) 2 + CO2 = CaCO3 + H2O – ion exchange, calcium carbonate precipitate is formed;

We make calculations:
M Ca (OH) 2 = 74 g / mol;

M (CaCO3) = 100 g / mol;

Y Ca (OH) 2 = m / M = 200/74 = 2.7 mol (substance in excess).

Proportion:
1 mol of gas at normal level – 22.4 liters;

X mol (CO2) – 33.6 liters from here, X mol (CO2) = 1 * 33.6 / 22.4 = 1.5 mol;

Y (CO2) = 1.5 mol (deficient substance);

Y (CaCO3) = 1.5 mol since the amount of substances is 1 mol.

Calculations are made for the substance in deficiency.

Find the mass of salt:
m (CaCO3) = Y * M = 1.5 * 100 = 150 g

Answer: The mass of calcium carbonate is 150 g