Balance the equation
`Mg+ HNO_(3) to Mg(NO_(3))_(2) + N_(2)O + H_(2)O`

Explanation

Step Find out the elemnts which undergo a change in oxidation nukber (O.N)

Here O.N of Mg increases form 0 in Mg metal to +2 in `Mg(NO_(3))_(2)` and htat of N decreases form +5 `HNO_(3) "to" +1 "in" N_(2)O`
Step 2 Find oiut the total increase and decrease in O.N
Since there is only one Mg atom on either side of Eq (i) therefore total increase in O.N of Mg is 2 Further since there are two N atoms in `N_(2)O` on R.H.S and only one in `HNO_(3)` on L.H.S of Eq (i) therefore multiple `HNO_(3)` on L.H.S of Eq (i) by 2 and thus the total decrease in O.N of N is `2xx4=8`
Step 3 Balance increse/decrease in O.N
sine the total incrase in O.n is 2 and decrease is 8 therefore multiply Mg on the L.H.S of Eq (i) by combining steps 2 and 3 above we hve
step 4 Balance all atoms other than O and H
To balance Mg on either side of Eq (ii) multiply `Mg(NO_(3))_(2)(aq) + N_(2)O(g)+H_(2)O(l)`
Now there are 10 nitrogne atoms on rhs of equy (iii) nd only 2 on L.H.S therefroe to balance N atom change the coeffiecient of `HNO_(3)` form 2 ot 10 on L.H.S of Eq (iii) we have
`4 Mg (s) +10 HNO_(3) (aq)` to `4 Mg (NO_(3))_(2)(aq)+N_(2)O(g)+H_(2)O(l)`
Step 5 Balance O and H atoms by hit and trial method since there are 30 oxygen atoms on L.H.S but only 26 oxygen atoms on R.H.S of Eq (iv) therefore to balacne O atoms change hte coeffiecent of `H_(2)O` from 1 to 5 we have
`4 Mg (S) + 10 HNO_(3)(Aq) to Mg (BO_(3))_(2)(aq)+N_(2)O(g)+5H_(2)O(l)`
the H atoms get automatically balaced Thus Eq (v) repersents hte correct balanced equation