Question

I am so confused with my chemistry lab right now. We were given the following equations to work with:

[Equation 1] 2 I^- + H2O2 + 2 H^+ ----> I2 + 2 H2O
[Equation 9] 2S2O3^2- + I2 ----> 2I^- +S4O6^2-

And During the Experiment we used the following for each reaction:
T1 | 10 mL of 0.3 M KI | 10mL of 0.02 M Na2S2O3 | 30 mL of 0.1 M H2O2
T2 | 20 mL of 0.3 M KI | 10mL of 0.02 M Na2S2O3 | 30 mL of 0.1 M H2O2
T3 | 30 mL of 0.3 M KI | 10mL of 0.02 M Na2S2O3 | 30 mL of 0.1 M H2O2
T4 | 30 mL of 0.3 M KI | 10mL of 0.02 M Na2S2O3 | 50 mL of 0.1 M H2O2
T5 | 30 mL of 0.3 M KI | 10mL of 0.02 M Na2S2O3 | 70 mL of 0.1 M H2O2

For the lab we also used water, a buffer, and starch in the solutions.

(From lab manual - Calculate the number of moles of S2O3^2- that is consumed in each reaction. Use stoichiometry in equation 9. From equation 9 one-half the moles of S2O3^2- equals the moles of I2 that form in the reaction. Record the moles of I2 formed during the elapsed time)

I have no idea where to start... So, to calculate the number of moles of S2O3^2-, would I have to take 0.02M Na2S2O3 and multiply it by the .010L (10 mL) that we used for the solutions to find the moles - which would be 2 x 10^-4 mol Na2S2O3. But how would I find the amount consumed. Would all of it be consumed in the end? Because in this lab, once all the S2O3^2- is consumed the solution turns dark blue. And how do I find I2. Some help would be really appreciated.

Answer 1

All of the numbers run together with no spaces so it's hard to know what you've done. However, equation 1 is a reaction that FORMS I2. To know how much is formed you titrate it with standard S2O3^2-. Yes, mols S2O3^2- = M x L and mols I2 which came from equation 1 will be 1/2 the thiosulfate mols. You titrate all of the I2 with thiosulfate and that's when the starch turns blue. Again, that's the amount of I2 formed from the first reaction.