Question

Some molecules of `C_11H_24` crack to produce ethene and propene. How would you write a balanced equation to show a cracking reaction which makes ethene and propene?

Answer 1

`C_11H_24 -> C_2H_4 + C_3H_6 + C_6H_14`

Explanation:

The starting molecule in the question, undecane, will break down into three products. Two are listed in your question: ethene and propene (these are alkenes). The third product is an alkane that consists of the "leftover" carbons and hydrogens. Ethene has the formula `C_2H_4`. Propene has the formula `C_3H_6`.

So far, you have:

`C_11H_24 -> C_2H_4 + C_3H_6 + C_?H_?`

Remember that we have to balance the equation, so the leftover carbons and hydrogens will go to the third product. We have `11-2-3 = 6` carbons and `24-4-6=14` hydrogens. This forms the alkane known as hexane.

Now the balanced equation is:

`C_11H_24 -> C_2H_4 + C_3H_6 + C_6H_14`

Cracking involves larger alkanes, so hexane should not break down.