Answer to Question #35637 in Chemistry for sousou

Question

At 1.00 bar 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.

Answer 1

First, let's determine the total number of moles of both gases in the mixture at given conditions via Mendeleev-Clapeyron's law: PV=nRT; n=(PV)/(RT)= (101.3*10^3*5.04*10^-3)/(8.31*273)=0.225 mol.
Then, let's write down to equations for combustion of gases:
CH4+2O2=CO2+2H2O;
C3H8+5O2=3CO2+4H2O;
Let's solve the system of 2 equations with 2 unknown. We indicate number of moles of methane as x, and number of moles of propane as y.
x+y=0.225
44x+132y=16.5
b=0.075; a=0.15
Thus, the molar fraction of methane would be 0.15/0.225=0.(6) and the fraction of propane is 0.(3)
The mixture of initial gas consists of 66.6% of methane and 33.3% of propane