# Having trouble balancing a chemical equation. How would I balance this?

## $N {H}_{3}$ + ${O}_{2}$ $\rightarrow$ NO + ${H}_{2} O$ I find that I have 1 N on both sides, 3 H's on the left and 2 H's on the right. The O's are balanced. I can balance the H's (giving them a total of 6 and both sides), but I can't seem to figure out how to do with the others, without throwing the whole thing off.

Nov 12, 2015

Here's how you can go about balancing this chemical equation.

#### Explanation:

This chemical equation is a little tricky to balance if you don't want to use oxidation numbers, which is assume is the case here.

The first thing to look for in such cases is to balance the elements that are present in odd number on one side of the equation and in even number on the other side of the equation.

In this case, you have $3$ atoms of hydrogen on the reactants' side and $2$ on the products' side. To balance these out, multiply the ammonia molecule by $2$ and the water molecule by $3$

$\textcolor{red}{2} \text{NH"_3 + "O"_2 -> "NO" + color(green)(3)"H"_2"O}$

Now, the nitrogen and the oxygen atoms are now unbalanced. Focus on the nitrogen atoms first, and leave the oxygen atoms for last, you'll see why in a minute.

Since you now have $2$ atoms of nitrogen on the reactants' side, and only $1$ on the products' side, multiply the nitric oxide molecule by $2$ to get

$\textcolor{red}{2} \text{NH"_3 + "O"_2 -> color(blue)(2)"NO" + color(green)(3)"H"_2"O}$

Now everything is balanced except the oxygen atoms, since you have $5$ on the products' side and only $2$ on the reactants' side.

On the reactants' side, oxygen is present as a molecule, ${\text{O}}_{2}$, which means that you have some elbow room in terms of what stoichiometric coefficient you can assign here.

So, what number multiplied by $2$ will give you $5$?

$x \cdot 2 = 5 \implies x = \frac{5}{2}$

This means that if you multiply the oxygen molecule by $\frac{5}{2}$, you will have $5$ oxygen atoms on both sides of the equation.

$\textcolor{red}{2} \text{NH"_3 + 5/2"O"_2 -> color(blue)(2)"NO" + color(green)(3)"H"_2"O}$

Finally, to remove the fractional coefficient, multiply all the species by $2$

$\textcolor{red}{4} \text{NH"_3 + 5"O"_2 -> color(blue)(4)"NO" + color(green)(6)"H"_2"O}$

And there you have it, the balanced chemical equation for this reaction.