Find the volume of carbon monoxide (IV), which will be released by the action of hydrochloric acid on 22 g of limestone

Find the volume of carbon monoxide (IV), which will be released by the action of hydrochloric acid on 22 g of limestone containing 5% non-carbonate impurities. Perform calculations with an accuracy of hundredths.

Given:
w (impurities) = 5%
m (mixture) = 22g
Find: V (CO2)
Decision:
CaCO3 + 2HCl = CaCl2 + H2O + CO2
w (pure) (CaCO3) = 100% – w (impurities) = 95%
m (pure) (CaCO3) = w (pure) (CaCO3) * m (mixture) = 95% * 22g / 100% = 20.9g
n (CaCO3) = m (CaCO3) / M (CaCO3) = 20.9 g / 100g / mol = 0.209 mol
And the reaction equation implies that n (CaCO3) = n (CO2)
n (CO2) = 0.209 mol
V (CO2) = Vm * n (CO2) = 22.4 L / mol * 0.209 mol = 4.68 L
Answer: 4.68 liters