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

##### 2 Answers

#### Explanation:

There is no fixed method actually. You have to take a random number as the number of molecules for

As for here, I have taken

Now, hydrogen atoms are present only in

Now, when we calculated the number of molecules of

On the other hand, as there are

That makes

So now the total number of oxygen atoms is

So the final balanced reaction is :

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

Two possible answers: (1)

#### Explanation:

First, you need to tally all the atoms.

*Left side:*

N = 1

H = 3

O = 2

*Right side:*

N = 1

H = 2

O = 2 + 1 (two from the **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

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

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

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

*Right side:*

N = 1 x 2 = **2**

H = 2 x 3 = **6**

O = (2 x 2) + (1 x 3) = **7**

Hence,

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

** #(cancel 2)#** [

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