Answer:
[tex]M_f=38.8\%[/tex]
Explanation:
From the question we are told that:
Pressure [tex]P=747mmHg[/tex]
Temperature [tex]T=298K[/tex]
Volume [tex]V=11.1[/tex]
Heat Produced [tex]Q=780kJ[/tex]
Generally the equation for ideal gas is mathematically given by
 [tex]PV=nRT[/tex]
 [tex]n= (747/760) *11.1/ (0.0821*298)[/tex]
 [tex]n=0.446mol[/tex]
Therefore
 [tex]x+y=0.446[/tex]
 [tex]x=0.446-y .....1[/tex]
Since
Heat of combustion of Methane=889 kJ/mol
Heat of combustion of Propane=2220 kJ/mol
Therefore
 [tex]x(889) + y(2220) = 760 ...... 2[/tex]
Comparing Equation 1 and 2 and solving simultaneously
 [tex]x=0.446-y .....1[/tex]
 [tex]x(889) + y(2220) = 760 ...... 2[/tex]
 [tex]x=0.173[/tex]
 [tex]y=0.273[/tex]
Therefore
Mole fraction 0f Methane is mathematically given as
 [tex]M_f=\frac{x}{n}*100\%[/tex]
 [tex]M_f=\frac{1.173}{0.446}*100\%[/tex]
 [tex]M_f=38.8\%[/tex]
