J.R. S. answered 04/02/18
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This is an oxidation/reduction (redox) reaction. It isn't quite as straight forward as some other reactions you may have been dealing with. In this reaction, the oxidation half reaction involves oxidation of Cl- to Cl2 gas and reduction half reaction is the reduction of MnO4- to Mn2+.
Oxidation: 2Cl- → Cl2 + 2e-
Reduction: 8H+ + MnO4- + 5e- → Mn2+ + 4H2O
In order to balance the electrons in the ox and red reactions, we multiply oxidation by 5 and reduction by 2 and then add the two reactions together and cancel where appropriate. We get...
16H+ + 2MnO4- + 10Cl- → 2Mn2+ + 8H2O + 5Cl2
Then filling in the spectator ions (if you must), you get
2KMnO4 + 16HCl → 2MnCl2 + 5Cl2 + 2KCl + 8H2O
Then filling in the spectator ions (if you must), you get
2KMnO4 + 16HCl → 2MnCl2 + 5Cl2 + 2KCl + 8H2O