Question

A gaseous fuel mixture stored at 736 mm Hg and 298 K contains only methane (CH₄) and propane (C₃H₈). When 11.0 L of this fuel mixture is burned, it produces 722 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

Temperature of the stored gaseous fuel mixture = T = 298 K

Pressure of the stored gaseous fuel mixture = P = 736 mm of Hg

now 760 mm of Hg is equivalent to 1 atm, i.e., 760 mm Hg = 1 atm

Lets find out the number of moles of gas present in the mixture

The combustion energies of methane and propane are 889 kJ/mol and 2200 kJ/mol respectively

Let the number of moles of methane in mixture to be x and that of propane to be y

also total number of moles of methane and propane is 0.435 moles

therefore solving above two equations we get

and

The mole fraction of methane in the mixture is given by

Therefore the mole fraction of methane in the given mixture is 41.14%