Answer
$2CH_{3}$$CH_{2}$$CH_{2}$$CH_{3}$ + $13O_{2}$ --> $8CO_{2}$ + $10H_{2}O$
Work Step by Step
Hydrocarbons undergo a combustion reaction, Carbon dioxide and water are given off as the end products of the reaction.
If we try to balance this equation directly, we get 13 oxygen atoms on the right-hand side, which means we need to put 6.5 in front of oxygen. 6.5 $\times$2=13 .
Therefore to get a whole number we balance it by adding 2 in front of the alkane. thus, now
1. There are 8 Carbon atoms on the left-hand side which gets balanced when we balance carbon dioxide with an 8.
$2CH_{3}$$CH_{2}$$CH_{2}$$CH_{3}$ --> $8CO_{2}$
2. There are 20 hydrogen atoms on the left-hand side which gets balanced when we place 10 in front of a water molecule. (10$\times$2= 20)
$2CH_{3}$$CH_{2}$$CH_{2}$$CH_{3}$ --> $10H_{2}O$
3. There are 26 oxygen atoms on the right-hand side. 8$\times$2= 16+10=26
These get balanced when we place 13 in front of an oxygen molecule on the left-hand side. (13$\times$2=26)
$2CH_{3}$$CH_{2}$$CH_{2}$$CH_{3}$ + $13O_{2}$ --> $8CO_{2}$ + $10H_{2}O$
