At 1.00 atm and 0 °c, a 5.04 l mixture of methane (ch4) and propane (c3h8) was burned, producing 16.5 g of co2. what was the mole fraction of each gas in the mixture? assume complete combustion.

The chemical reactions involved are:

CH4 + 2 O2 → CO2 + 2 H2O
C3H8 + 5 O2 → 3 CO2 + 4 H2O

Moles of gases = 5.04 L ( 1 mol / 22.41 L ) = 0.22 mol gases

Moles C in CO2 produced = 15.0 g CO2 ( 1 mol CO2 / 44.01 g CO2 ) (1 mol C/1 mol CO2)= 0.34 mol C

Let x be the molar fraction of CH4 in the mixture so 1-x is the molar fraction of C3H8 in the mixture.

We set up a C balance:

0.34083 mol C = (0.22 mol)(x) (1 mol C / 1 mol CH4) + (0.22 mol) (1-x) (3 mol C / 1 mol C3H8)  

x = 0.74 methane
1-x = 0.26 propane

The answer would be