Through lime passed 1 liter of carbon monoxide (2) and carbon monoxide (4). The resulting precipitate was filtered

Through lime passed 1 liter of carbon monoxide (2) and carbon monoxide (4). The resulting precipitate was filtered off, washed and dried. Its mass was found to be 2.45 g. Determine the volume fraction of each gas in the initial mixture.

Given:
V mixture (CO, CO2) = 1 l
m (sediment) = 2.45 g
Vm = 22.4 l / mol

To find:
φ (CO) -?
φ (CO2) -?

Decision:
1) Ca (OH) 2 + CO2 => CaCO3 ↓ + H2O;
2) n (CaCO3) = m (CaCO3) / M (CaCO3) = 2.45 / 100 = 0.0245 mol;
3) n (CO2) = n (CaCO3) = 0.0245 mol;
4) V (CO2) = n (CO2) * Vm = 0.0245 * 22.4 = 0.5488 l;
5) φ (CO2) = V (CO2) * 100% / V mixture (CO, CO2) = 0.5488 * 100% / 1 = 54.88%;
6) φ (CO) = 100% – φ (CO2) = 100% – 54.88% = 45.12%.

Answer: The volume fraction of CO is 45.12%; CO2 – 54.88%.