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

Oct 23, 2015

$4 N {H}_{3} + 7 {O}_{2} \to 4 N {O}_{2} + 6 {H}_{2} O$

#### Explanation:

There is no fixed method actually. You have to take a random number as the number of molecules for $N {H}_{3}$, then calculate result.

As for here, I have taken $4$ as the number of molecules of $N {H}_{3}$. That means $4$ nitrogen atoms and $12$ hydrogen atoms are taking part in this reaction.

Now, hydrogen atoms are present only in ${H}_{2} O$ as a product in this reaction. So that makes $6$ molecules of ${H}_{2} O$ (as there are $12$ hydrogen atoms).

Now, when we calculated the number of molecules of ${H}_{2} O$, we also got the number of oxygen atoms present there.

On the other hand, as there are $4$ molecules of $N {H}_{3}$, there has to be $4$ molecules of $N {O}_{2}$ 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 $\left(8 + 6\right)$ or $14$, which means $14$ atoms or $7$ molecules of oxygen.

So the final balanced reaction is :

$4 N {H}_{3} + 7 {O}_{2} \to 4 N {O}_{2} + 6 {H}_{2} O$

Oct 24, 2015

Two possible answers: (1) $2 N {H}_{3}$ + $\frac{7}{2} {O}_{2}$ = $2 N {O}_{2}$ + $3 {H}_{2} O$ or (2) $4 N {H}_{3}$ + $7 {O}_{2}$ = $4 N {O}_{2}$ + $6 {H}_{2} O$

#### Explanation:

First, you need to tally all the atoms.

$N {H}_{3}$ + ${O}_{2}$ = $N {O}_{2}$ + ${H}_{2} O$

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

Right side:
N = 1
H = 2
O = 2 + 1 (two from the $N {O}_{2}$ and one from ${H}_{2} O$ ; 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 ${H}_{2} O$

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

$2 N {H}_{3}$ + ${O}_{2}$ = $N {O}_{2}$ + $3 {H}_{2} O$

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

$2 N {H}_{3}$ + ${O}_{2}$ = $2 N {O}_{2}$ + $3 {H}_{2} O$

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 $\frac{7}{2}$)

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

Hence,

$2 N {H}_{3}$ + $\frac{7}{2} {O}_{2}$ = $2 N {O}_{2}$ + $3 {H}_{2} O$

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

$\left(\cancel{2}\right)$ [$2 N {H}_{3}$ + $\frac{7}{\cancel{2}} {O}_{2}$ = $2 N {O}_{2}$ + $3 {H}_{2} O$]

=

$4 N {H}_{3}$ + $7 {O}_{2}$ = $4 N {O}_{2}$ + $6 {H}_{2} O$