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