# How do you balance this equation? ?H_2O_2 -> ?H_2O + ?O_2

Mar 12, 2017

$2 {H}_{2} {O}_{2} \left(l\right) \to 2 {H}_{2} O \left(l\right) + {O}_{2} \left(g\right)$

#### Explanation:

Given:

?H_2O_2 -> ?H_2O+?O_2

Since we only have one molecule on the left, we can start using fractions:

${H}_{2} {O}_{2} \to {H}_{2} O + \frac{1}{2} {O}_{2}$

Since the water molecule on the right of the equation contains the same number of hydrogens as the hydrogen peroxide on the left, its multiplier must be the same.

This leaves one oxygen atom - i.e. $\frac{1}{2} {O}_{2}$ unaccounted for.

To give use integral multipliers, double all the quantities.

$2 {H}_{2} {O}_{2} \to 2 {H}_{2} O + {O}_{2}$

Finally we can annotate with the probable phases: liquid for water and hydrogen peroxide, gas for oxygen.

$2 {H}_{2} {O}_{2} \left(l\right) \to 2 {H}_{2} O \left(l\right) + {O}_{2} \left(g\right)$