Mole fraction of Methane In Mixture

A gaseous fuel mixture stored at 737 mm Hg and 298 K contains only methane (CH4) and propane (C3H8). When 13.8 L of this fuel mixture is burned, it produces 906 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.)


Any help would be greatly appreciated! :)

Enthalpy of combustion of methane = -890 kJ/mol propane = - 2219 kJ/mol (check with your data).

Number of moles of gas in mixture, n = PV/RT so find n

Let number of moles of methane = w moles, propane = n-w.

w x -890 + (n-w)x -2219 = -906 . Find w. Mole fraction CH4 = w/n

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