There is no fixed an approach actually. You need to take a random number together the variety of molecules because that #NH_3#, then calculation result.

As for here, I have taken #4# as the number of molecules of #NH_3#. That means #4# nitrogen atoms and also #12# hydrogen atoms room taking component in this reaction.

You are watching: Nh3+o2=no2+h2o

Now, hydrogen atoms are existing only in #H_2O# as a product in this reaction. For this reason that provides #6# molecule of #H_2O# (as there room #12# hydrogen atoms).

Now, once we calculate the variety of molecules the #H_2O#, we also got the number of oxygen atoms existing there.

On the various other hand, together there space #4# molecule of #NH_3#, there needs to be #4# molecules of #NO_2# as well to balance the variety of nitrogen atom in both side of the reaction.

That renders #8# an additional atoms of oxygen.

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

So the final well balanced reaction is :

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


Answer attach
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Nikka C.
Oct 24, 2015

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


Explanation:

First, you should 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 native #H_2O# ; DO NOT include IT up YET )

You constantly have to find the simplest aspect that you deserve to balance (in this case, the H).

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

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

Notice that through balancing the atoms, you perform 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 must multiply the right side N through 2.

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

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

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

Now, the only aspect left come be well balanced is O. Because 7 is one odd number, you have the right to use a portion for the equation to be in its reduced form.

Left side:N = 1 x 2 = 2H = 3 x 2 = 6O = 2 x 3.5 = 7 (the decimal 3.5 have the right to be composed as #7/2#)

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

Hence,

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

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