Determine the volume of gas that is formed by the interaction of 100 g of calcium arbonate

Determine the volume of gas that is formed by the interaction of 100 g of calcium arbonate, in which 10% of impurities with hydrochloric acid

According to the condition of the problem, we write down the equation of the process:
CaCO3 + 2HCl = CaCl2 + CO2 + H2O – ion exchange, carbon dioxide is released;

Calculations by formulas:
M (CaCO3) = 100 g / mol;

M (CO2) = 44 g / mol.

Determine the mass, the amount of the original substance:
100 * (1 – 0.10) = 90 g (mass of calcium carbonate salt without impurities);

Y (CaCO3) = m / M = 90/100 = 0.9 mol;

Y (CO2) = 0.9 mol since the amount of substances is 1 mol.

Find the volume of the product:
V (CO2) = 0.9 * 22.4 = 20.16 L

Answer: received carbon monoxide (4) with a volume of 20.16 liters