The balanced equation doesnt seem to help because the thiosulfate is not in the equation, I think it is preventing the reaction from going backwards. THe equation is:

2iodide ion + hydrogen peroxide + 2 H3O --> elemental iodide + 4 H2O
within the reaction there is also deionized water, an acetate buffer, 0.3 M KI, 0.02 M Na2S2O3 (Sodium thiosulfate) and starch. and then hydrogen peroxide is added to that solution.
It says the thiosulfate removes the elemental iodide as it is formed so I think that is preventing the reaction from going backwards.
I don't know the amount of the product and am just given the amount of each solution, because it is a rate problem, I am supposed to determine the rate experimentally but I don't understand how you can find the amount of thiosulfate reacted when I am only given sodium thiosulfate.
Is it possible to separate the sodium from the sodium thiosulfate? I think the sodium from the sodium thiosulfate reacts with the buffer as it becomes a acetic acid-sodium acetate buffer.
I am assuming that to get this I must use either the rate law equation or the activation energy equation but those don't seem to help any.

How can you find the amount of moles reacted in a rate problem. I am given the molarity of sodium thiosulfate ( 0.002 M) and I am to find the amount of thiosulfate ion reacted in moles. I also have the temperature and the length of the reaction. Can I separate the thiosulfate ion from the sodium thisulfate with stoichiometry?
Any help is appreciated

I would look to the balanced equation. Do you know the amount of product?

Thiosulfate removes the elemental iodine as follows:
S2O3^-2 + I2 ==> S4O6^-2 + 2I^-
Yes, you may separate thiosulfate ion from sodium thiosulfate.
Na2S2O3(s) + H2O ==> 2Na^+(aq) + S2O3^-2(aq)

You are correct that the acetate is there to buffer the solution but your reasoning is not correct. The original equation you wrote contains H3O^+. That is the ion being controlled by the NaC2H3O2 (sodium acetate) and not the sodium ion.
You can calculate the amount of thiosulfate used by knowing how much I2 was formed. I hope this wlll get you started.

how can you find out how much I2 is formed
I think that that is the second step after I find the amount of thiosulfate reacted.
I know the molarity of KI and Na2S2O3 and the volume of each of those solutions so I can get the moles or grams of those but how can I get them when they are a combination of 2 elements?

You don't have enough information in all of the posts for us to know exactly what you are doing or how you are doing it. So we are trying to answer questions piece meal. Have you balanced the equation? I think it should be
2H^+ + H2O2+ 2I^- ==> I2 + 2H2O

As soon as the I2 is formed it is reduced back to I^- by thiosulfate. When does the starch indicator turn? What time are you measuring? Is the H2O2 standardized? If not, is anything standardized? If the H2O2 has been standardized, that is what is used to determine the amount of I2 released.

The starch indicator turns blue at various times because there are various amounts of each solution. The H2O2 is standardized in the first couple of trials so I guess that I can use that

Yes, if the H2O2 concentration is known then I2 will be known.

ok thanks i think i got it

So what is the rate reaction?

The rate of reaction can be determined by measuring the change in concentration of a reactant or product over time. In this case, since you are given the molarity of sodium thiosulfate (Na2S2O3) and are trying to find the amount of thiosulfate ion reacted in moles, you can use the known concentration and stoichiometry of the balanced equation to determine the rate.

First, make sure the balanced equation is properly written:

2H+ + H2O2 + 2I- -> I2 + 2H2O

Next, determine the stoichiometric ratio between thiosulfate ion (S2O3^-2) and iodine (I2) in the reaction:

S2O3^-2 + I2 -> S4O6^-2 + 2I^-

From the balanced equation, you can see that 1 mole of thiosulfate ion reacts with 1 mole of iodine to produce 1 mole of tetrathionate ion (S4O6^-2) and release 2 moles of iodide ions (I^-).

To calculate the amount of thiosulfate reacted, you will need to know the molarity and volume of the sodium thiosulfate solution used in the reaction. Multiply the molarity by the volume to get the number of moles of thiosulfate ion present in the solution.

Once you have the moles of thiosulfate ion reacted, you can use the stoichiometric ratio to determine the moles of iodine produced. Since the stoichiometry is 1:1 between thiosulfate and iodine, the moles of iodine will be the same as the moles of thiosulfate reacted.

Please note that without additional information, such as the concentrations of other reactants or the rate constant of the reaction, it may not be possible to determine the rate of the reaction experimentally using just the given data.