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Chile salt peter, a source of $\ce{NaNO3}$ also contains $\ce{NaIO3}$. The $\ce{NaIO3}$ can be used as a source of iodine, produced in the following reactions.

\begin{align} \ce{IO3- + 3 HSO3- &-> I- + 3 H+ + 3 SO4^2-} \tag{i}\\ \ce{5 I- + IO3- + 6 H+ &-> 3 I2 (s) + 3 H2O} \tag{ii} \end{align}

One litre of chile salt peter solution containing $\pu{5.80 g}$ $\ce{NaIO3}$ is treated with stoichiometric quantity of $\ce{NaHSO3}$. Now an additional amount of same solution is added to the reaction mixture to bring about the second reaction. How many grams of $\ce{NaHSO3}$ are required in step (i) and what additional volume of chile salt peter must be added in step (ii) to bring in complete conversion of $\ce{I-}$ to $\ce{I2}$?

My working out using Mole concept: enter image description here

My working out using equivalence: enter image description here

Why does the equivalence concept not work? Please tell me what is wrong and what is the correct method to use equivalence.

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    $\begingroup$ Just a note: capitalization matters in chemistry. Io – Jupiter's moon, whereas IO – iodine monooxide. $\endgroup$ – andselisk♦ Oct 14 '17 at 14:18
  • $\begingroup$ Sorry I find typing really hard hence the pictures $\endgroup$ – TheLostGuardian0 Oct 14 '17 at 14:20
  • $\begingroup$ I did an OCR on the photo from the textbook and it would be nice if you actually proceed with typing your attempt using MD and MathJax as well. In the last sentence, you ask whether eq. of $\ce{NaIO3}$ = eq. of $\ce{NaIO3}$, which seems to be a typo. $\endgroup$ – andselisk♦ Oct 14 '17 at 15:09
  • $\begingroup$ I edited it and re wrote it tell me if it's better $\endgroup$ – TheLostGuardian0 Oct 14 '17 at 16:31

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