Answer
$2C_5H_{10}O_2(l)+13O_2(g)\rightarrow 10CO_2(g)+10H_2O(g)$
Work Step by Step
$C_5H_{10}O_2(l)+O_2(g)\rightarrow CO_2(g)+H_2O(g)$
Add a coefficient of 5 to $CO_2$ to balance $C$ and to $H_2O$ to balance $H$.
$C_5H_{10}O_2(l)+O_2(g)\rightarrow 5CO_2(g)+5H_2O(g)$
This gives an odd number of $O$ atoms on the right, which can never be balanced on the left. Double all coefficients to eliminate the odd number.
$2C_5H_{10}O_2(l)+2O_2(g)\rightarrow 10CO_2(g)+10H_2O(g)$
Now all atoms are in balance except $O$. There are 8 $O$ on the left and 30 on the right, so a shortage of 22 on the left. Increase the coefficient of $O_2$ from 2 to 13 to add the needed 22 atoms.
$2C_5H_{10}O_2(l)+13O_2(g)\rightarrow 10CO_2(g)+10H_2O(g)$
