Answer
$CH_{3}$$CH_{2}$$CH=CH_{2}$ + $6O_{2}$ --> $4CO_{2}$ + $4H_{2}O$
Work Step by Step
Hydrocarbons undergo a combustion reaction, Carbon dioxide and water are given off as the end products of the reaction.
1. There are 4 Carbon atoms on the left-hand side which gets balanced when we balance carbon dioxide with a 4.
$CH_{3}$$CH_{2}$$CH=CH_{2}$ --> $4CO_{2}$
2. There are 8 hydrogen atoms on the left-hand side which gets balanced when we place 4 in front of a water molecule. (2$\times$4= 8)
$CH_{3}$$CH_{2}$$CH=CH_{2}$ --> $4H_{2}O$
3. There are 12 oxygen atoms on the right-hand side. (4$\times$2= 8+4=12)
These get balanced when we place 6 in front of an oxygen molecule on the left-hand side. (6$\times$2=12)
$CH_{3}$$CH_{2}$$CH=CH_{2}$ + $6O_{2}$ --> $4CO_{2}$ + $4H_{2}O$
