Question #35637

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.

Expert's answer

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

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

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