What volume of carbon dioxide is formed during the combustion of 72 g of methane, if the mass fraction

What volume of carbon dioxide is formed during the combustion of 72 g of methane, if the mass fraction of carbon dioxide output was 69%.

1. Let’s write the reaction equation:
CH4 + 2O2 = CO2 + 2H2O.
2. Find the amount of methane:
n (CH4) = m (CH4) / M (CH4) = 72 g / 16g / mol = 4.5 mol.
3. According to the reaction equation, we find the theoretical amount of carbon monoxide, and at the end its volume (Vm is the molar volume, constant equal to 22.4 l / mol):
ntheor (CO2) = n (CH4) = 4.5 mol.
nprak (CO2) = ntheor (CO2) * η (CO2) / 100% = 4.5 mol * 69% / 100% = 3.105 mol.
V (CO2) = nprak (CO2) * Vm = 3.105 mol * 22.4 l / mol = 69.6 l.
Answer: V (CO2) = 69.6 liters.