KIO3 + NaHSO3 ---> I2 + Na2SO4 + H2SO4 + K2SO4 + H20.
The balanced equation is 4 KIO3 + 10 NaHSO3 --> 2 I2 + 5 Na2SO4 + 3 H2SO4 + 2 K2SO4 + 2 H2O.
Is this the answer to mine, but I just have to add starch to both sides?
There didn't appear to be a precipitate in our mixture and how am I supposed to know that H2O is a product.... Can you explain this to me?
Well, as I mentioned before, the reactions here are a little difficult at first glance. You should know that nobody would anticipate the reactions.... That is the students next to you don't intuitively know the iodate plus bisulfite gives iodide and sulfate.
First, the equation you posted is correct. At a cursory glance your balanced eq looks ok.
The starch issue can be solved often by writing it as "starch" on the left, and then "starch*I2 complex" on the right.... Though with the 2 coefficient in front of I2, if they want you to include starch, you'd have to say 2 starch ---> 2 complex... Not sure if they'd care or not...
There is no precipitate. Not sure why you might think there would be one... But the complex that formed is the dark color you observed
You wouldn't intuitively know water was formed, and there's no visual cues. Again, the reaction steps are not intuitive. I'm surprised your lab manual didn't cover the reaction steps. I taught with a manual that we made, and we always included the reactions in the prelab discussion.
Here... This breaks down the reation.... Its more than you need for the overall reaction, but it might help explain some of the things that are happening:
First, the iodate reacts with the bisulfite:
1) KIO3 + 3 NaHSO3 ---> KI + 3 NaHSO4
Which can also be written as
IO3- + 3 HSO3- ---> I- + 3 SO42- + 3 H+
Then, some of the formed iodide reacts with more iodate to get iodine (I2)
2) 5 I- + IO3- + 6 H+ ---> 3 I2 + 3 H2O
But, the iodine will not react with starch as long as bisulfite is present. The reaction with starch is slower than the reaction with bisulfite. So the iodine will react with bisulfite to form more sulfate, until there is no more bisulfite.
3) I2 + HSO3- + H2O ---> 2 I- + SO42- + 3 H+
When there is no more bisulfite, then the I2 produced in step 2 will then complex with the starch.
4) I2 + starch ---> iodine-starch complex (blue-black)
Over all, if you look back at the reactions, you can see that some of the I- produced in 1) is used in 2)... But then more I- is made again in 3). Also, iodate and bisulfate are consumed in 1) and 2) or 3)... So it gets used up. And then, as mentioned, I2 is both made and consumed in 2) and 3).... But more must get made than consumed in 3), otherwise you'd never get reaction with starch 4).
As I said. This is not a simple reaction. It has several rxns, all of which are not intuitive, and can be intimidating.
But look at the steps, see where things cancel out, and the overall rxn makes more sense.