# How do you balance the following equation: ?NH_3 + ?O_2 -> ?NO + ?H_2O?

Jan 21, 2017

Overall,

$4 N {H}_{3} + 5 {O}_{2} \rightarrow 4 N O + 6 {H}_{2} O$

#### Explanation:

These are standard redox reactions, nitrogen is oxidized, and oxygen is reduced:

$\text{Oxidation:}$

$N {H}_{3} + {H}_{2} O \rightarrow N O + 5 {H}^{+} + 5 {e}^{-}$ $\left(i\right)$

$\text{Reduction:}$

${O}_{2} + 4 {H}^{+} + 4 {e}^{-} \rightarrow 2 {H}_{2} O$ $\left(i i\right)$

We cross multiply and take $4 \times \left(i\right) + 5 \times \left(i i\right)$

$4 N {H}_{3} + 5 {O}_{2} + 4 {H}_{2} O \rightarrow 4 N O + 10 {H}_{2} O$

Nitrogens, oxygens, and hydrogens all balance. Charges balance, so this is a valid representation. Of course we can substract 4 equiv of water from EACH side of the equation.

Ammonia might be fully oxidized up to $N {O}_{2}$. If you represent this reaction for practice, would you post the balanced equation in this thread?