How do you balance __ NH3 + __O2--> __NO2 + __H2O? The number in the parenthasis are subscribts, thank you all so much!!!?

There is no fixed method actually. You have to take a random number as the number of molecules for `NH_3`, then calculate result.

As for here, I have taken `4` as the number of molecules of `NH_3`. That means `4` nitrogen atoms and `12` hydrogen atoms are taking part in this reaction.

Now, hydrogen atoms are present only in `H_2O` as a product in this reaction. So that makes `6` molecules of `H_2O` (as there are `12` hydrogen atoms).

Now, when we calculated the number of molecules of `H_2O`, we also got the number of oxygen atoms present there.

On the other hand, as there are `4` molecules of `NH_3`, there has to be `4` molecules of `NO_2` as well to balance the number of nitrogen atoms in both side of the reaction.

That makes `8` another atoms of oxygen.

So now the total number of oxygen atoms is `(8+6)` or `14`, which means `14` atoms or `7` molecules of oxygen.

So the final balanced reaction is :

`4NH_3 + 7O_2 -> 4NO_2 + 6H_2O`

First, you need to tally all the atoms.

`NH_3` + `O_2` = `NO_2` + `H_2O`

Left side:
N = 1
H = 3
O = 2

Right side:
N = 1
H = 2
O = 2 + 1 (two from the `NO_2` and one from `H_2O` ; DO NOT ADD IT UP YET )

You always have to find the simplest element that you can balance (in this case, the H).

Left side:
N = 1 x 2 = 2
H = 3 x 2 = 6
O = 2

Right side:
N = 1
H = 2 x 3 = 6
O = 2 + (1 x 3) from the O in `H_2O`

Notice that with balancing the atoms, you do not forget that they are part of a substance - meaning, you have to multiply everything.

`2NH_3` + `O_2` = `NO_2` + `3H_2O`

Now, the N is not balanced so you need to multiply the right side N by 2.

Left side:
N = 1 x 2 = 2
H = 3 x 2 = 6
O = 2

Right side:
N = 1 x 2 = 2
H = 2 x 3 = 6
O = (2 x 2) + (1 x 3) = 7

`2NH_3` + `O_2` = `2NO_2` + `3H_2O`

Now, the only element left to be balanced is O. Since 7 is an odd number, you can use a fraction for the equation to be in its reduced form.

Left side:
N = 1 x 2 = 2
H = 3 x 2 = 6
O = 2 x 3.5 = 7 (the decimal 3.5 can be written as `7/2`)

Right side:
N = 1 x 2 = 2
H = 2 x 3 = 6
O = (2 x 2) + (1 x 3) = 7

Hence,

`2NH_3` + `7/2O_2` = `2NO_2` + `3H_2O`

but if you want the equation to show whole numbers only, you can always multiply everything by the denominator in the fraction.

`(cancel 2)` [`2NH_3` + `7/(cancel 2)O_2` = `2NO_2` + `3H_2O`]

=

`4NH_3` + `7O_2` = `4NO_2` + `6H_2O`