Question

The products obtained by cracking an alkane, X, are methane, ethene and propene.

The mole fraction of ethene in the products is 0.5.

What is the identity of X?
A C6H14
B C8H18
C C9H20
D C11H24

Answer is B. I don't really understand the question, what does it mean the mole fraction of ether is 0.5?

Answer 1

ok lets write out the rxn:

CxHy . . . = . . . a CH4 . . . + . . . b C2H4 . . . + . . . c C3H6
1 mol x % mol. 50 % mol y%mol

here we know that x+y will equal 50% or antoher way of saying it is

n(C2H4) = n(CH4) + n(C3H8)


A. C6H14 = CH4 + C2H4 + C3H6 from above 1=1+1 no
D C11H24 = CH4 + .2 C2H4 +2 C3H6 2=1+2 no

try b and c