# Question #0f9d6

Always balance the most complicated molecules first (we'll leave ${O}_{2}$ last here). We see that we have 1 carbon on both sides, so that's fine, but we have 4 H in the reactants and only 2 H in the products. Make the coefficient of ${H}_{2} O$ 2 and then the hydrogens will be balanced. Now all that's left to do is balance O_2, which is simple. We see that we now have 4O on the products side, since we added the 2 in front of ${H}_{2} O$ and there is another 2O in $C {O}_{2}$. All we have to do is add a 2 in front of the ${O}_{2}$.