How many mmol of KIO3 is this?

Question

A student crushed a vitamin C tablet that weighed 1.634g, and transferred 1.552g of the powder into a 100 mL volumetric flask. After diluting to the 100.00mL mark, they titrated a 20.00mL aliquot of this solution against 0.0106 M KIO3 ( the KIO2 was in the buret) with an excess HCl and KI. The initial buret reading was 0.53 mL; the final buret reading was 17.29 mL.

Answer 1

KIO₃ + 5I^- + 6H⁺ ==> 3I₂ + K⁺ + 3H₂O

I₂ + C₆H₈O₆ ==> C₆H₆O₆ + 2I^- + 2H⁺

How many mmoles KIO3?

volume KIO₃ used = 17.29 ml - 0.53 ml = 16.76 ml

mols KIO₃ used = 16.76 ml x 1 L / 1000 ml x 0.0106 mol / L = 1.777x10^-4 mols

mmoles KIO₃ used = 1.777x10^-4 mols x 1 mmol / 1x10^-3 mol = 0.1777 mmoles KIO₃