Gọi x là mol C2H6, y là mol C3H8
C2H6+ 7/2O2 (t*)-> 2CO2+ 3H2O
C3H8+ 5O2 (t*)-> 3CO2+ 4H2O
=> nCO2= 2x+3y mol; nH2O= 3x+4y mol
nCO2: nH2O= 11:15
=>(2x+3y)/(3x+4y)= 11/15
<=> 11(3x+4y)= 15(2x+3y)
<=> 3x= y
<=> x/y= 1/3
nC2H6= x => nC3H8= 3x
%C2H6= x.100: (x+3x)= 25%