Answer
$2CH_{2}$$=CHCH_{3}$ + $9O_{2}$ --> $6CO_{2}$ + $6H_{2}O$
Work Step by Step
Hydrocarbons undergo a combustion reaction, Carbon dioxide and water are given off as the end products of the reaction.
If we try to balance this equation directly, we get 9 oxygen atoms on the right-hand side, which means we need to put 4.5 in front of oxygen. 4.5 $\times$2=9
Therefore to get a whole number we balance it by adding 2 in front of the alkene. thus, now
1. There are 6 Carbon atoms on the left-hand side which gets balanced when we balance carbon dioxide with a 6.
$2CH_{2}$$=CHCH_{3}$ --> $6CO_{2}$
2. There are 12 hydrogen atoms on the left-hand side which gets balanced when we place 6 in front of a water molecule. (6$\times$2= 12)
$2CH_{2}$$=CHCH_{3}$ --> $6H_{2}O$
3. There are 18 oxygen atoms on the right-hand side. 6$\times$2= 12+6=18
These get balanced when we place 9 in front of an oxygen molecule on the left-hand side. (9$\times$2=18)
$2CH_{2}$$=CHCH_{3}$ + $9O_{2}$ --> $6CO_{2}$ + $6H_{2}O$.
