Determine the volume of carbon monoxide (IV) released during the interaction of 250 grams

Determine the volume of carbon monoxide (IV) released during the interaction of 250 grams of calcium carbonate with hydrochloric acid.

1. Let’s write the equation of interaction of calcium carbonate with hydrochloric acid:

CaCO3 + 2HCl = CaCl2 + H2O + CO2 ↑;

2.Calculate the chemical amounts of calcium carbonate and released carbon dioxide:

n (CaCO3) = m (CaCO3): M (CaCO3);

M (CaCO3) = 40 + 12 + 3 * 16 = 100 g / mol;

n (CaCO3) = 250: 100 = 2.5 mol;

n (CO2) = n (CaCO3) = 2.5 mol;

3.determine the volume of carbon monoxide (IV):

V (CO2) = n (CO2) * Vm = 2.5 * 22.4 = 56 liters.

Answer: 56 l.