What volume of carbon dioxide will be released during the decomposition of 50 g of chalk, which contains 20% of impurities?

We write down the equation of reactions.
CaCO3 → CO2 + CaO.
First, we find the mass of impurities contained in the chalk.
m = 50 g × 0.2 = 10 g impurities.
Next, we find the mass of pure chalk.
m (CaCO3) = 50 g – 10 g = 40 g.
Next, we find the amount of chalk substance.
n = m / M.
M (CaCO3) = 40 + 12 + 3 × 16 = 100 g / mol.
n = 40 g / 100 g / mol = 0.4 mol CaCO3.
Next, we find the amount of carbon dioxide substance according to the reaction equation.
0.4 mol CaCO3 – x mol CO2.
1 mol of CaCO3 – 1 mol of CO2.
Find the unknown value of x.
X = 0.4 × 1 ÷ 1 = 0.4 mol of CO2.
Next, we find the volume of carbon dioxide.
n = V / Vm.
V = n × Vm = 0.4 mol × 22.4 mol = 8.96 L.
Answer: the volume of carbon dioxide is 8.96 liters.