Answer
$2C_5H_{12}O(l)+15O_2(g)\rightarrow 10CO_2(g)+12H_2O(l)$
Work Step by Step
When a hydrocarbon burns in air the other reactant is oxygen.
The basic equation for the reaction is:
$C_5H_{12}O(l)+O_2(g)\rightarrow CO_2(g)+H_2O(l)$
Add a coefficient of 5 to $CO_2$ to balance $C$.
Add a coefficient of 6 to $H_2O$ to balance $H$.
$C_5H_{12}O(l)+O_2(g)\rightarrow 5CO_2(g)+6H_2O(l)$
The count of $O$ atoms is odd on the left and even on the right. Multiply all coefficients by 2 to eliminate this situation.
$2C_5H_{12}O(l)+2O_2(g)\rightarrow 10CO_2(g)+12H_2O(l)$
There are 32 atoms of $O$ on the right. There are 6 on the left. Increase the coefficient of $O_2$ from 2 to 15 to balance $O$.
$2C_5H_{12}O(l)+15O_2(g)\rightarrow 10CO_2(g)+12H_2O(l)$
