When potassium permanganate reacts with hydrochloric acid forms potassium chloride, manganese chloride, chlorine and water

This is a great redox issue including MnO4-. The way to it is the ionic type of the redox condition for this particle which is:

8 H+ + MnO4-+ 5 e-→ Mn2+ + 4 H2O

Since we are shaping atomic Cl2 , we require another redox ionic condition

2 Cl-→ Cl2 + 2 e-

Since the e-must counterbalance when we consolidate these conditions, we have to duplicate the first by 2

what's more, various the second one by 5 and after that consolidate to get

16 H+ + 2 MnO4-+ 10 Cl-→ 2 Mn2+ + 8 H2O + 5 CL2

This is the crucial step. The rest is simply an issue of adding the two K+ particles to every side and six more Cl-particles to every side to deal with the MnCl2 and KCl that will be on the correct hand side.

The last outcome is:

2 K MnO4 + 16 HCl → 2 MnCl2 + 5 Cl2 + 2 KCl + 8 H2O