Question

A gaseous fuel mixture stored at 752mmHg and 298K contains only methane (CH4) and propane (C3H8). When 11.2 L of this fuel mixture is burned, it produces 765 KJ ofheat.

Answer 1

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 beformed 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.

Reworking the numbers in the problem above, we have
(802.5 kJ/mol) (X) + (2043.85 kJ/mol) (0.5165 - X) = 761 kJ of heat

802.5X + 1055.65 – 2043.85X = 761

1055.65 - 761 = 2043.85X – 802.5X

294.65 = 1241.35X

X = 0.2374 moles of Methane out of a total of 0.5165 moles,

For a mole fraction of 45.96%

Answer 2

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 balancedequations.
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.

Here is a similar result:

Thermochemistry Challenge:
"A gaseous fuel mixture stored at 748mmHg and 298K contains only methane, CH4, and propane, C3H8. When 12.9 L of this fuel mixture is burned, it produces761 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.)"

Method I (incorrect)

Using the ideal gas law, find moles:
PV = nRT so n = PV/RT
n = [(744/760 atm) (12.9 L)] / [(.08206)(298K)]
n= .978947*12.9/24.45388
n = 0.5165 total moles of gas

Look up heats of combustion
methane : 890 kJ/mol
propane : 2,220 kJ/mol

Let X = moles of Methane, then (0.5165 - X ) = moles of propane
Then
(890 kJ/mol) (X) + (2,220 kJ/mol) (0.5165 - X) = 761 kJ of heat
Or
890X + 1146.55 - 2220X = 761
And
1146.55 - 761 = 2220X - 890X
Finally
385.55 = 1330X
Or
X = 0.2899 moles of Methane
Thus the mole fraction of methane in the mixture would be
(0.2899 moles of Methane / 0.5165 total moles of gas) times 100% = 56.12 %
56.1 % moles are methane

This answer was rejected, so we decided to try another approach.

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 beformed 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.

Reworking the numbers in the problem above, we have
(802.5 kJ/mol) (X) + (2043.85 kJ/mol) (0.5165 - X) = 761 kJ of heat

802.5X + 1055.65 – 2043.85X = 761

1055.65 - 761 = 2043.85X – 802.5X

294.65 = 1241.35X

X = 0.2374 moles of Methane out of a total of 0.5165 moles,

For a mole fraction of 45.96%

Answer 3

Heat of combustion of methane = 889 kJ/mol
Heat of combustion of propane = 2,220 kJ/mol
Let moles of methane be x and moles of propane be y
(x+ y) * R * 298 = 11.2 * 752/760
=> x+y = 0.0372/R = 0.453
Also
889x + 2220y = 765
889 x + 2220 ( 0.453-x) = 765
=> 889x -2220x + 1005.66 = 765
=> -1331x = -240.66
=> x = 0.1808
Hence Mole fraction of methane = x/(x+y) = 0.1808/0.453 = 0.3991
Hence percentage mole fraction = 0.3991 * 100 = 39.91 %
Hope this helps. :)