What is the mass of sediment formed by the interaction of 5.6 liters of carbon dioxide through lime water?

To solve, we compose the equation of the process:

CO2 + Ca (OH) 2 = CaCO3 + H2O – ion exchange, calcium carbonate precipitate is obtained;
Calculations:
M (CO2) = 44 g / mol;

M (CaCO3) = 100 g / mol.

3. Proportion:

1 mole of gas at normal level – 22.4 liters;

X mol (CO2) -5.6 L from here, X mol (CO2) = 1 * 5.6 / 22.4 = 0.25 mol;

Y (CaCO3) = 0.25 mol since the amount of substances is 1 mol according to the equation.

4. Find the mass of the product:

m (CaCO3) = Y * M = 0.25 * 100 = 25 g

Answer: The mass of calcium carbonate is 25 g