Chemistry: The Central Science (13th Edition)

$2C_3H_7OH (l) + 9O_2 (g) \rightarrow 6CO_2 (g) + 8H_2O (l)$
In every combustion reaction, $CO_2$ and water products. The balanced equation is provided in the answer, but an effective method to balanced chemical equations would be to count the number of molecules for each element on both sides of the reaction. Only change the coefficient and not the subscript. We start with the given unbalanced form: $C_3H_7OH (l) + O_2 (g) \rightarrow CO_2 (g) + H_2O (l)$ Left: C:3, H:8,O:3 Right: C:1,H2,O:3 $C_3H_7OH (l) + O_2 (g) \rightarrow 3CO_2 (g) + H_2O (l)$ Left: C:3, H:8,O:3 Right: C:3,H:2,O:7 $C_3H_7OH (l) + O_2 (g) \rightarrow 3CO_2 (g) + 4H_2O (l)$ Left: C:3, H:8,O:3 Right: C:3,H:8,O:7 $C_3H_7OH (l) + 2O_2 (g) \rightarrow 3CO_2 (g) + 4H_2O (l)$ Left: C:3, H:8,O:5 Right: C:3,H:8,O:7 $2C_3H_7OH (l) + 2O_2 (g) \rightarrow 3CO_2 (g) + 4H_2O (l)$ Left: C:6, H:16,O:6 Right: C:3,H:8,O:6 $2C_3H_7OH (l) + 2O_2 (g) \rightarrow 6CO_2 (g) + 4H_2O (l)$ Left: C:6, H:16,O:6 Right: C:6,H:8,O:13 $2C_3H_7OH (l) + 2O_2 (g) \rightarrow 6CO_2 (g) + 8H_2O (l)$ Left: C:6, H:16,O:6 Right: C:6,H:16,O:20 $2C_3H_7OH (l) + 9O_2 (g) \rightarrow 6CO_2 (g) + 8H_2O (l)$ Left: C:6, H:16,O:20 Right: C:6,H:16,O:20