Question

A gaseous fuel mixture stored at 744 mm Hg and 298 K contains only methane (CH4) and propane (C3H8). When 13.1 L of this fuel mixture is burned, it produces 858 kJ of heat. What is the mole fraction of methane in the mixture? (Assume that the water produced by the combustion is in the gaseous state.)

Answer 1

Step 1. Use PV=nRT to find total number of moles of the gas mixture. Be sure to use the proper units for R.
Step 2. Find the heats of reaction for methane and propane by looking up the heats of formation of reactants and products using the balanced equations.
Step 3. Let the moles of methane be X and the moles of propane be total moles-X.
Step 4. Solve for mole fraction as a percent.

Using the ideal gas law, find moles:
PV = nRT so n = PV/RT
n = [(744/760 atm) (13.1 L)] / [(.08206)(298K)]
n = 0.5244 total moles of gas

In order to use the evolved heat, we might be required to use the balanced equation somehow, rather than merely the heats of combustion.

We know that for every Methane (CH4) mole burned, one CO2 and two H2O's must form in gaseous state. Similarly for every propane burned, there must be formed three CO2's and four H2O's. So we can use the heats evolved for these partial reactions to solve for the total heat evolved.

For example
CH4 + 2 O2 → CO2 + 2 H2O

Therefore, we know from our prior work or text that
Heat of formation of methane CH4(g) -74.6
Heat of formation of propane C3H8 -103.85
Heat of formation of CO2(g) -393.5
Heat of formation of H2O(g) -241.8

So for the balanced reaction of combustion of methane, we find
(-393.5 – 2*241.8)+74.6 = -802.5.
And similarly, for propane,
3*-393.5 – 4*241.8 + 103.85 = -2043.85.

These are NOT the standard molar heats of combustion in reference tables.

working the numbers in the problem above, we have
(802.5 kJ/mol) (X) + (2043.85 kJ/mol) (0.5244 - X) = 858 kJ of heat

802.5X + 1071.79 – 2043.85X = 858

1071.79 - 858 = 2043.85X – 802.5X

213.79 = 1241.35X

X = 0.1722 moles of Methane out of a total of 0.5244 moles,

Thus the mole fraction of methane in the mixture would be
(0.1722 moles of Methane / 0.5244 total moles of gas) times 100% =

For a mole fraction of 32.84%     -----------------------          your answer.