Balance the following equation by oxidation number method
`Zn+HNO_(3) to Zn(NO_(3))_(2)+NH_(4)NO_(3)+H_(2)`

Explanation

Writing oxidation number of al the atoms.
`overset(0)(Zn)+overset(+1 +5 -2)(HNO_(3))to overset(+2 +5 -2)(Zn(NO_(3))_(2))+overset(-3 +1)(NH_(4))overset(+5 -2)(NO_(3))+overset(+1)(H_(2))overset(-2)(O) `
The oxidation number of Zn and N have changed.
`overset(0)(Zn) to overset(+2)(Zn)(NO_(3))_(2) ...(i) `
`Hoverset(+5)(NO_(3)) to overset(-3)(NH_(4))NO_(3) ...(ii) `
Inrease in Ox. no. of Zn=2 unit per Zn atom
Decrease in Ox. no. of N=8 units per `HNO_(3)` molecule
Eq. (i) should be multiplied by 4 and added to eq. (ii)
`4Zn+HNO_(3)to 4Zn(NO_(3))_(2)+NH_(4)NO_(3)`
To balance nitrogen, 9 molecule of `HNO_(3)` should be added on LHS.
To balance hydrogen and oxygen, `3H_(2)O` molecules should be addd on RHS. Hence, the balanced equation is
`4Zn+10HNO_(3) to 4Zn(NO_(3))_(2)+NH_(4)NO_(3)+3H_(2)O`