Answer: The balanced equation is 4 NH3 + 7 O2 --> 4 NO2 + 6 H2O
Step-by-step explanation:
Oxidation numbers
We begin by noting the oxidation numbers on each atom:
NH3: N = -3, H = +1
O2: O = 0
-----------------
NO2: N = +4, O = -2
H2O: H = +1, O = -2
From this we conclude that N has lost 7 electrons, while O has gained 2.
Half-reactions
Because this case is more complex, we must consider the half-reactions:
Eq. 1) N^-3 --> N^+4 + 7 e- (which says that N has lost 7 electrons)
Eq. 2) O + 2 e- --> O^-2 (which says that O has gained 2 electrons)
However, oxygen is a diatomic atom, so we must adjust the second equation by a factor of 2.
Eq. 1) N^-3 --> N^+4 + 7 e-
Eq. 2) O2 + 4 e- --> 2 O^-2
Balancing the Half-reactions
To balance the half-reactions, we must multiply each equation by a factor that allows us to cancel out the electrons. The LCM of 7 and 4 is 28, so we multiply the first equation by 4, and the second equation by 7, giving:
Eq. 1) 4 N^-3 --> 4 N^+4 + 28 e-
Eq. 2) 7 O2 + 28 e- --> 14 O^-2
Now, adding them together gives:
4 N^-3 + 7 O2 --> 4N^+4 + 14 O^-2
All together...
To complete the reaction, we must add the missing hydrogen atoms. With 4 nitrogen atoms on the left, we require 12 hydrogen atoms to construct the 4 NH3 molecules. We will add 12 H to both sides:
4 N^-3 + 7 O2 --> 4N^+4 + 14 O^-2
+ 12 H + 12 H
-------------------------------------------------------
4 N^-3 + 12 H + 7 O2 --> 4N^+4 + 14 O^-2 + 12 H
We can now combine all the atoms together into the known molecules on each side:
4 NH3 + 7 O2 --> 4N^+4 + 8 O^-2 + 12 H + 6 O^-2
which is,
4 NH3 + 7 O2 --> 4 NO2 + 6 H2O