Balance the following equation. (for a balanced eq. aA + bB → cC + dD, enter your answer as the integer abcd)

MnO4−(aq) + SO32−(aq) + H+(aq) → Mn2+(aq) + SO42−(aq) + H2O(l)

5 answers

I have tried putting in diff answers but its not working
This is a redox, you have to balance charge. Have you studied redox?

origSpecies NewSpecies change
Mn+7.............Mn+2 .......gained 5 electrons

S+4...............S+6.......lost two electrons

SO,you will need to Mn for each five S

2MnO4−(aq) + 2.5(SO3)2−(aq) + H+(aq) → Mn2+(aq) + SO42−(aq) + H2O(l)
is the first go. Then start balancing mass then get rid of the fraction..
4MnO4−(aq) + 5(SO3)2−(aq) + H+(aq) → 2Mn2+(aq) + 5(SO4)2−(aq) + H2O(l)
then balance the O (dont mess with any S related O, so balance the water O
4MnO4−(aq) + 5(SO3)2−(aq) + H+(aq) → 2Mn2+(aq) + 5(SO4)2−(aq) + H2O(l)
I see on the left 16+30 O on the left, on the right 40 0 on the right (not cournting the water O), so finally then
4MnO4−(aq) + 5(SO3)2−(aq) + H+(aq) → 2Mn2+(aq) + 5(SO4)2−(aq) + 6H2O(l)
that balances the O
then finally, the H
4MnO4−(aq) + 5(SO3)2−(aq) + 6H+(aq) → 2Mn2+(aq) + 5(SO4)2−(aq) + 6H2O(l)

Now recheck it all.
For some reason its still not working
Forget it i figured it out
2MnO4{-}(aq)+5 SO3{2-}(aq)+6 H{+}(aq)→2 Mn{2+}(aq) + 5 SO4{2-}(aq) + 3 H2O(l)

2 I{-}(aq) + 4 H{+}(aq) + 2 NO2{-}(aq) → 2 NO(g) + 2 H2O(l) + I2(s)