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?

Aug 18, 2017

${C}_{11} {H}_{24} \to {C}_{2} {H}_{4} + {C}_{3} {H}_{6} + {C}_{6} {H}_{14}$

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 ${C}_{2} {H}_{4}$. Propene has the formula ${C}_{3} {H}_{6}$.

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:

${C}_{11} {H}_{24} \to {C}_{2} {H}_{4} + {C}_{3} {H}_{6} + {C}_{6} {H}_{14}$

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