When sodium carbonate interacts with hydrochloric acid, sodium chloride, water and 44.8 liters of carbon monoxide

When sodium carbonate interacts with hydrochloric acid, sodium chloride, water and 44.8 liters of carbon monoxide are formed (4). What amount of hydrochloric acid substance reacted?

1. Let’s compose the equation of the chemical reaction:

Na2CO3 + 2HCl = 2NaCl + CO2 + H2O.

2. Find the amount of carbon dioxide substance (Vm – molar volume, constant equal to 22.4 l / mol):

n (CO2) = V (CO2) / Vm = 44.8 l / 22.4 l / mol = 2 mol.

3. According to the equation of the chemical reaction, we find the amount of hydrochloric acid that has reacted:

n (HCl) = n (CO2) * 2 = 2 mol * 2 = 4 mol.

Answer: n (HCl) = 4 mol.