Answer
$CH_{3}$$CH_{2}$$CH_{3}$ + $5O_{2}$ --> $3CO_{2}$ + $4H_{2}O$
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 3 Carbon atoms on the left-hand side which gets balanced when we balance carbon dioxide with a 3.
$CH_{3}$$CH_{2}$$CH_{3}$ --> $3CO_{2}$
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_{3}$ --> $4H_{2}O$
3. There are 10 oxygen atoms on the right-hand side. (3$\times$2= 6 +4=10)
These get balanced when we place 5 in front of an oxygen molecule on the left-hand side. (5$\times$2=10)
$CH_{3}$$CH_{2}$$CH_{3}$ + $5O_{2}$ --> $3CO_{2}$ + $4H_{2}O$
