Find the volume of gas that was released when 250 g of calcium carbonate interacted with sulfuric acid.

1. Let’s write down the reaction equation:

CaCO3 + H2SO4 = CaSO4 + CO2 + H2O.

2. Find the amount of reacted calcium carbonate:

n (CaCO3) = m (CaCO3) / M (CaCO3) = 250 g / 100 g / mol = 2.5 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):

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

V (CO2) = n (CO2) * Vm = 2.5 mol * 22.4 l / mol = 56 l.

Answer: V (CO2) = 56 liters.