How do I balance this? Cr2(So4)3 +KI + KIO3 +H20=Cr (OH)3 +K2SO4 +I2?

Answer 1

To balance Cr2(SO4)3 +KI + KIO3 +H20=Cr (OH)3 +K2SO4 +I2
First, let us get rid of the spectators:
I(-) + IO3(-) = I2
So I(-) is oxidized from oxidation state of -1 to 0 and IO3(-) is reduced from oxidation state of +5 to 0:
5 I(-) = 2.5 I2 + 5e-........................(1)
IO3(-) + 3H2O + 5e- = 0.5 I2 + 6OH-....(2)
Add (1) and (2) together and cancel 5e-, we get:
5 I(-) + IO3(-) + 3H2O = 3 I2 + 6OH-
Now we put the spectators back to both sides (gradually):
2Cr(3+) + 5 KI + KIO3 + 3H2O = 3 I2 + 2Cr(OH)3 + 6K+
Cr2(SO4)3 + 5KI + KIO3 + 3H2O = 3 I2 + 2Cr(OH)3 + 3K2SO4
This is the requested balanced equation.