There is no fixed method actually. You have to take a random number together the variety of molecules for #NH_3#, then calculation result.

As for here, I have actually taken #4# together the variety of molecules that #NH_3#. That way #4# nitrogen atoms and #12# hydrogen atoms space taking component in this reaction.

You are watching: Nh3 + o2 = no + h2o

Now, hydrogen atoms are present only in #H_2O# together a product in this reaction. For this reason that makes #6# molecules of #H_2O# (as there room #12# hydrogen atoms).

Now, as soon as we calculate the number of molecules of #H_2O#, we additionally got the variety of oxygen atoms present there.

On the other hand, together there are #4# molecule of #NH_3#, there has to be #4# molecule of #NO_2# also to balance the number of nitrogen atoms in both side of the reaction.

That makes #8# another atoms the oxygen.

So currently the total variety of oxygen atoms is #(8+6)# or #14#, which way #14# atom or #7# molecule of oxygen.

So the final balanced reaction is :

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

Answer connect

Nikka C.

Oct 24, 2015

Two possible answers: (1) #2NH_3# + #7/2O_2# = #2NO_2# + #3H_2O# or (2) #4NH_3# + #7O_2# = #4NO_2# + #6H_2O#

Explanation:

First, you must tally all the atoms.

#NH_3# + #O_2# = #NO_2# + #H_2O#

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

*Right side:*N = 1H = 2O = 2 + 1 (two indigenous the #NO_2# and also one from #H_2O# ; **DO NOT include IT increase YET** )

You always have to find the simplest aspect that you have the right to 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 carry out not forget the they are part of a problem - meaning, you need to multiply everything.

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

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

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*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 come be balanced is O. Due to the fact that 7 is one odd number, you deserve to use a portion for the equation to be in its lessened form.

*Left side:*N = 1 x 2 = **2**H = 3 x 2 = **6**O = 2 x 3.5 = **7** (the decimal 3.5 deserve to be composed 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 desire the equation to show whole numbers only, you can constantly multiply whatever by the denominator in the fraction.