How many liters of carbon monoxide (4) can be obtained from limestone weighing 25 g with a mass

How many liters of carbon monoxide (4) can be obtained from limestone weighing 25 g with a mass fraction of impurities of 20%.

Given:
m (limestone) = 25 g
ω approx. = 20%

To find:
V (CO2) -?

1) CaCO3 => CaO + CO2 ↑;
2) ω (CaCO3) = 100% – ω approx. = 100% – 20% = 80%;
3) m (CaCO3) = ω (CaCO3) * m (limestone) / 100% = 80% * 25/100% = 20 g;
4) n (CaCO3) = m / M = 20/100 = 0.2 mol;
5) n (CO2) = n (CaCO3) = 0.2 mol;
6) V (CO2) = n * Vm = 0.2 * 22.4 = 4.5 liters.

Answer: The CO2 volume is 4.5 liters.