Find the volume of gas that is formed by the interaction of 106 g of carbonate with 36.5 g of hydrochloric acid.

1. Let’s write down the reaction equation:

CaCO3 + 2HCl = CaCl2 + CO2 + H2O.

2. Find the amount of calcium carbonate and hydrochloric acid

n (CaCO3) = m (CaCO3) / M (CaCO3) = 106 g / 100 g / mol = 1.06 mol.

n (HCl) = m (HCl) / M (HCl) = 36.5 g / 36.5 g / mol = 1 mol.

3. According to the reaction equation, we find the amount, and then the volume of released carbon dioxide (Vm – molar volume, constant equal to 22.4 l / mol):

! According to the reaction equation, it can be seen that hydrochloric acid is in short supply, therefore we solve it.

n (CO2) = n (HCl) / 2 = 1 mol / 2 = 0.5 mol.

V (CO2) = n (CO2) * Vm = 0.5 mol * 22.4 l / mol = 11.2 l.

Answer: V (CO2) = 11.2 liters.