To balance any redox equation, first divide the reaction into two half reactions, and balance the atom whose oxidation number is changing:
NO2 --> NH3
H2 --> H2O
Next, balance oxygens by adding H2O where needed, and then balance hydrogen by adding H+:
NO2 --> NH3 + 2 H2O
H2O + H2 --> H2O
7 H+ + NO2 --> NH3 + 2 H2O
H2O + H2 --> H2O + 2 H+
Next, balance charges by adding electrons (e-):
7 H+ + NO2 + 7 e- --> NH3 + 2 H2O
H2O + H2 --> H2O + 2 H+ + 2 e-
Now, multiply one or both equations by small numbers to make the number of electrons the same in the two half reactions. In this case, multiply the top equation by 2 and the bottom one by 7 to give you 14 electrons in each half reaction:
14 H+ + 2 NO2 + 14 e- --> 2 NH3 + 4 H2O
7 H2O + 7 H2 --> 7 H2O + 14 H+ + 14 e-
Finally, add the two half reactions and simplify by cancelling anything that is on both sides. This will give you the final balanced reaction:
2 NO2 + 7 H2 --> 2 NH3 + 4 H2O
So, the answer to the original question is (A) 7
