Question

Calculate the moles of IO₃^- dispensed if you have 21.56 mL KIO₃ M 1.0 and IO₃^- (aq) +8I^- (aq) +H⁺ (aq) -> 3I₃^- (aq) + 3H₂O(l). Then calculate the moles of I₃^- produced.

Then calculate the moles of C₆H₈O₆ in the sample. C₆H₈O₆ (aq) + I₃^- (aq) + H₂O(l) -> C₆H₈O₇(aq) + 3I^- (aq) + 2H⁺ (aq)

Thank you!

Answer 1

The overall reaction:

IO₃^- + 8I^- + H⁺ --------> 3I₃^- + 3H₂O

From there:

C₆H₈O₆ + I₃ + H₂O ---------> C₆H₈O₇ + 3I^- + 2H⁺

From your previous data given, we can calculate the moles of IO3:

Moles IO₃^- = 1 * 0.02156 = 0.02156 moles

According to your first reaction and assuming that the reaction is complete:

1 mol IO₃^- --------> 3I₃^-

0.02156 ----------> X

X = 0.02156 * 3 = 0.06468 moles

Now the relation in moles of I₃ and C₆H₈O₆ is the same, so the moles of I₃ are the same of C₆H₈O₆ which is 0.06468 moles.

Hope this helps