Answer
$1 C_4H_{10}O(l) + 6O_2(g) --> 4CO_2(g) + 5H_2O(l) $
Work Step by Step
1. Write the unbalanced equation:
$ C_4H_{10}O(l) + O_2(g) --> CO_2(g) + H_2O(l) $
* Remember that, the complete combustion of a molecule, is a reaction with $O_2$ to produce $CO_2$ and water.
2. Balance the number of Carbon moles:
$1 C_4H_{10}O(l) + O_2(g) --> 4CO_2(g) + H_2O(l) $
3. Balance the number of Hydrogen moles:
$1 C_4H_{10}O(l) + O_2(g) --> 4CO_2(g) + 5H_2O(l) $
4. Now we can count 13 Oxygen moles in the right side, and because the left side already has 1 with the $C_4H_{10}O(l)$, we can put a 6 next to the $O_2$ and the reaction will be completely balanced.
$1 C_4H_{10}O(l) + 6O_2(g) --> 4CO_2(g) + 5H_2O(l) $
