The product obtained by cracking an alkane x are methane ethene and propene. the mole fraction of ethene in the product is 0.5. what is the identity of x?

1) The alkanes have general formula CnH2n+2

For example

n= 1 => CH4    methane
n=2 => C2H6    ethane
n=3 => C3H8    propane
n=4 => C4H10    butane
n=8 => C8H18      nonane

2) The three alkenes produced are:

methane: CH4
ethene: C2H4
propene: C3H6

3) That the mole fraction of C2H4 is 0.5 means that the number of moles of C2H4 is the double of the other two.

Then If the other two have A mole each, the number of moles of C2H4 is 2A

That yields to the general reaction is:

CnH(2n+2) → A CH4 + 2A C2H4 + A C3H6

4) Now balance the C and the H

Balance on C: n = A + 4A + 3A = 8A => A = n/8

Balance on H: 2n + 2 = 4A + 8A + 6A = 18 A

=> 2n + 2 = 18 (n/8)

=> 18n/8 - 2n = 2

=> [18n - 16n] / 8 = 2

=> 2n = 16 => n = 16/2

=> n = 8 => A = 1

Then the alkane is C8H18 and the general reaction is:

C8H18 → CH4 + 2C2H4 + C3H6

C8H18

Sorry i dont understand the question can you rephase it?

The correct answer is "hydrogen".