Here is example 4b:

KMnO_{4}+ KCl + H_{2}SO_{4}---> K_{2}SO_{4}+ MnSO_{4}+ Cl_{2}+ H_{2}O

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 KMnO_{4}AND the KCl)

H_{2}SO_{4}

K_{2}SO_{4}

SO_{4}^{2}¯ (from the MnSO_{4})

H_{2}O

That leaves this:

MnO_{4}¯ + Cl¯ ---> Mn^{2+}+ Cl_{2}

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):

MnO3) add back in the spectator ions (which I will do in a step-wise fashion):_{4}¯ ---> Mn^{2+}

Cl¯ ---> Cl_{2}5e¯ + 8H

^{+}+ MnO_{4}¯ ---> Mn^{2+}+ 4H_{2}O

2Cl¯ ---> Cl_{2}+ 2e¯16H

^{+}+ 2MnO_{4}¯ + 10Cl¯ ---> 2Mn^{2+}+ 5Cl_{2}+ 8H_{2}O (this is the balanced, net ionic equation)

a) add back in 8 sulfates:

8H_{2}SO_{4}+ 2MnO_{4}¯ + 10Cl¯ ---> 2MnSO_{4}+ 5Cl_{2}+ 8H_{2}O + 6K_{2}SO_{4}b) This added 12 potassium ions to the right side, so add 12 K

^{+}to the left:8H_{2}SO_{4}+ 2KMnO_{4}+ 10KCl ---> 2MnSO_{4}+ 5Cl_{2}+ 8H_{2}O + 6K_{2}SO_{4}

The last equation is the balanced, molecular equation.