Here is example 4b:

KMnO4 + KCl + H2SO4 ---> K2SO4 + MnSO4 + Cl2 + H2O

I'm going to answer this problem in two different ways. I'll link to the second way at the end of this file.

1) eliminate all spectator ions (this means anything that isn't reduced or oxidized). The list is:

K+ (from the KMnO4 AND the KCl)
SO42¯ (from the MnSO4)

That leaves this:

MnO4¯ + Cl¯ ---> Mn2+ + Cl2

Why did I keep those? It's because they are the exact items oxidized (the Cl) or reduced (the Mn).

2) balance the net ionic reaction (in acidic solution because of the sulfuric acid):

MnO4¯ ---> Mn2+
Cl¯ ---> Cl2

5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O
2Cl¯ ---> Cl2 + 2e¯

16H+ + 2MnO4¯ + 10Cl¯ ---> 2Mn2+ + 5Cl2 + 8H2O (this is the balanced, net ionic equation)

3) add back in the spectator ions (which I will do in a step-wise fashion):

a) add back in 8 sulfates:

8H2SO4 + 2MnO4¯ + 10Cl¯ ---> 2MnSO4 + 5Cl2 + 8H2O + 6K2SO4

b) This added 12 potassium ions to the right side, so add 12 K+ to the left:

8H2SO4 + 2KMnO4 + 10KCl ---> 2MnSO4 + 5Cl2 + 8H2O + 6K2SO4

The last equation is the balanced, molecular equation.

Here is the second solution