## Chemistry: The Central Science (13th Edition)

$1CH_3OH(l) + \frac{3}{2}O_2(g) --> 1CO_2(g) + 2H_2O(g)$
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.