How do you balance MnO4- + S (2-) --> MnS + S?

this is a basic redox reaction... so you add H2O to the product side and H+ to the reactant side... then balance the charges... the hint said you could add S-2 to the product side, but i'm not sure how to balance this|||Easier to use oxidation states. Then you get 2, 5, 2, 3.



Then 16H+ and 8H2O.|||When you balance an equation it means that everything is equal on both sides, now it's been a LONG time since I've balanced an equation so make sure you check this work. Since you said it is a redox reaction and you add H+ to the reactant side and H20 to the product side I'll rewrite it that way:


MnO4-+S(2-)+H---%26gt;MnS+SO2+H2O





This way you still have 1 Mn, 2 Ss, and 4 0s, plus your H.