Answer
$1C_8H_{18}(l) + \frac{25}{2} O_2(g) --> 8 CO_2(g) + 9 H_2O(l)$
Work Step by Step
1. Write the unbalanced equation:
*Remember that, in a complete combustion, the molecule reacts with $O_2$ to produce $CO_2$ and $H_2O$.
$C_8H_{18}(l) + O_2(g) --> CO_2(g) + H_2O(l)$
2. First, balance the number of Carbon atoms, by multiplying the number of moles of $CO_2$ by 8
$C_8H_{18}(l) + O_2(g) --> 8 CO_2(g) + H_2O(l)$
3. Now balance the number of Hydrogen atoms, by multiplying the number of moles of $H_2O$ by 9.
$C_8H_{18}(l) + O_2(g) --> 8 CO_2(g) + 9 H_2O(l)$
4. Finally, we can count 25 Oxygen atoms in the right side, so we can multiply the number of moles of $O_2$ by $\frac{25}{2}$.
$C_8H_{18}(l) + \frac{25}{2}O_2(g) --> 8 CO_2(g) + 9 H_2O(l)$
