Balance the following reactions by oxidation number method
`KMnO_(4) + H_(2)SO_(4) +HCl rarr K_(2) SO_(4) + MnSO_(4) + H_(2) O + Cl_(2)`

Explanation

Oxidation number method:
Follow the step given below to balance the reaction given above.
Assign oxidation state to the atoms that the oxidised or reduced. Note that oxidation state of `O,K,H` and `S` is same on both sides.
`KMoverset(+7)nO_(4)+H_(2)SO_(4)+K_(2)overset(+3)C_(2)O_(4)rarroverset(+2)MnSO_(4)+overset(+4)CO_(2)`
Write two half reactions and balance the atoms as follows:
a. `Mn^(7+)rarrMn^(2+)`("reduction")
b. `C_(2)^(3+)rarr2C^(4+)("oxidation")`
Balance the charge by adding `5e^(-)` to left of (a) and `2e^(-)` to the right of (b).
a. `Mn^(7+)+5e^(-)rarrMn^(2+)`(reduction)
b. `C_(2)^(3+)rarr2C^(4+)+2e^(-)`(oxidation)
Add two half reactions after multiplying (a) by `2` and (b) by `5` in order that electrons from both sides cancel each other.
`2Mn^(7+)+5C_(2)^(3-)rarr2Mn^(2+)+10C^(4+)`
Now compare this balanced equation with the molecular unbalanced equation as follows:
`2KMnO_(4)+5K_(2)C_(2)O_(4)rarr2MnSO_(4)+10CO_(2)`
As the charges balanced, now balance `K` atoms on both sides by adding `6K_(2)SO_(2)` on the rignt.
`2KMnO_(4)+5K_(2)C_(2)O_(4)rarr2MnSO_(4)+10CO_(2)+6K_(2)SO_(4)`
Now to balance `SO_(4)^(2-)` ions on both sides, and `8H_(2)SO_(4)` on the left.
`2KMnO_(4)+5K_(2)C_(2)O_(4)+8H_(2)SO_(4)rarr2MnSO_(4)+10CO_(2)+6K_(2)SO_(4)`
Finally add `8H_(2)O` on the right to balance. `O` and `H` atoms to get balanced equation.
`2KMnO_(4)+5K_(2)C_(2)O_(4)+8H_(2)SO_(4)rarr2MnSO_(4)+10CO_(2)+6K_(2)SO_(4)+8H_(2)O`
Note: Make a final check by counting `O` atoms on both sides.