Answer
$1CH_3OH(l) + \frac{3}{2}O_2(g) --> 1CO_2(g) + 2H_2O(g)$
Work Step by Step
1. Write the unbalanced equation:
$CH_3OH(l) + O_2(g) --> CO_2(g) + H_2O(g)$
2. Balance the number of Carbon moles:
$1CH_3OH(l) + O_2(g) --> 1CO_2(g) + H_2O(g)$
3. Balance the number of Hydrogen moles:
$1CH_3OH(l) + O_2(g) --> 1CO_2(g) + 2H_2O(g)$
4. Balance the number of Oxygen moles:
$1CH_3OH(l) + \frac{3}{2}O_2(g) --> 1CO_2(g) + 2H_2O(g)$
*Since we got 4 Oxygen moles in the right side, and we already have 1 in $1CH_3OH(l)$, we can put a $(\frac{3}{2})$ next to the $O_2$ and the equation will be totally balanced.