A gaseous fuel mixture stored at 750 mmHg and 298 K contains

A gaseous fuel mixture stored at 750 mmHg and 298 K contains only methane CH4 and propane (C3H8). When 11.0 L of this fuel mixture is burned, it produces 761 Kj of heat.<?xml:namespace prefix = o ns = \"urn:schemas-microsoft-com:office:office\" />

What is the mole fraction of methane in the mixture? (Assume that the water produced by the combustion is in the gaseous state.)

Solution

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

Using the ideal gas law, find moles:

PV = nRT so n = PV/RT
n = [(750/760 atm) (11.0L)] / [(.08206)(298K)]
n= 0.9868*11.0/24.45388
n =0.444 total moles of gas

So for the balanced reaction of combustion of methane, we find
(-393.5