Thank you for providing the additional information.
1 mol I3^- is consumed by 2 mols S2O3^2- so moles S2O3^2-/2 = mols I3^-
If that 0.0013 mols I3^- is the amount of I3^- formed from the addition of IO3^-, then 1/2 mols S2O3^2- will give you the excess I3^- formed. Hope this helps.
How do you calculate the moles of triiodide that is consumed by a thiosulfate titrant in a Vitamin C titration?
IO3– + 8 I– + 6 H+ -> 3 I3– + 3 H2O
I3– + 2 S2O32– -> 3 I– + S4O62–
Moles of I3- = 0.0013
Endpoint = 11.05 ml
Please let me know if you need more info.
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