CBSE Class 11-science Answered

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

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 is 4 but there are two N atoms in N2O and one in HNO3.Hence, decrease in Oxidation number of N is 2*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 can be written as :

4 Mg(aq) +2 HNO3(aq) ---> Mg(NO3)2 (aq) + N20 (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+5 H2O

Here H atoms get balanced automatically .