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

##
#NH_3# + #O_2# #rarr# NO + #H_2O#

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.

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.

##### 1 Answer

#### Answer:

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

#color(red)(2)"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

#color(red)(2)"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

On the reactants' side, oxygen is present as a molecule,

So, what number multiplied by

#x * 2= 5 implies x= 5/2#

This means that if you multiply the oxygen molecule by

#color(red)(2)"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

#color(red)(4)"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.