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?

1 Answer
Aug 18, 2017

Answer:

#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.