CBSE Class 11-science Answered

1.       Find out the element which undergoes a change in oxidation number:

The oxidation number of Mg increases from 0 to +2 in Mg metal to Mg(NO₃)₂ and N decrease from +5 in HNO₃ to +1 in N₂O

 2.       Find out the total increase/decrease in oxidation number :

Total Increase in Mg is 2 and decrease in N is 4 but there are two N atoms in N₂O and one in HNO₃. Hence, the decrease in Oxidation number of N is 2 X 4 =8

3.       Balance increase/decrease in Oxidation number:

Since the total increase is 2 and decrease  is 8.thereore, multiply Mg on LHS by 4 and combine step 2 and 3

Hence equation will be written as :

4 Mg(aq) +2 HNO₃(aq) → Mg(NO₃)₂ (aq) + N₂O (g)+ H₂O

4.       Balance all atoms other than O and H:

To balance Mg on either side, multiply (MgNO3) by 4 we have,

4 Mg +2HNO₃ → 4 Mg (NO₃)₂+N₂O+H₂O

Now there are 10 N atoms on RHS and 2 on LHS, therefore to balance N atoms change the coefficients of HNO3 from 2 to 10 on LHS and we have,

4 Mg(NO₃)₂+10HNO₃ → 4 Mg (NO₃)₂+N₂O+H₂O

5.       Balance O and H :

There are 30 O atoms on LHS but only 26 atoms on RHS, therefore, to balance O atoms, change the coefficients of H2O from 1 to 5. We have,

4 Mg(NO₃)₂+2HNO₃ → 4 Mg (NO₃)₂+N₂O+5H₂O

Here H atoms get balanced automatically.