We are going to do this by means of electronic balance:
\(N(IV)\Rightarrow\ N(V)+e^-|*4\\
2O(0)+4e^-\Rightarrow\ 2O^2^-\\
4NO_2+O_2+2H_2O\Rightarrow4HNO_3\)
If you hadn't learnt it yet, you would do it normally:
Count oxygens. To have even number of oxygen write "2" in fronf of H2O. Then to have equal hydrogen you have to add "4" in front of HNO3 and then NO2.
