How many grams of ${\mathrm{NaHSO}}_3$ will be required to react with one litre of ${\mathrm{NaIO}}_3$ solution containing 5.8g of  ${\mathrm{NaIO}}_3$ according to the reaction ?

${\mathrm{IO}}_3^{-}+3{\mathrm{HSO}}_3^{-}\stackrel{ }{\to }{\mathrm{I}}^{-}+3{\mathrm{H}}^{+}+3{\mathrm{SO}}_4^{2-}$

 $\begin{array}{l}\mathrm{Meq}.\hspace{0.17em}\hspace{0.17em}\mathrm{of}\hspace{0.17em}\hspace{0.17em}{\mathrm{NaHSO}}_3\mathrm{\hspace{0.17em}}=\mathrm{\hspace{0.17em}}\mathrm{Meq}.\hspace{0.17em}\hspace{0.17em}\mathrm{of}\hspace{0.17em}\hspace{0.17em}{\mathrm{NaIO}}_3\\ \mathrm{N}\hspace{0.17em}\hspace{0.17em}\times \hspace{0.17em}\hspace{0.17em}\mathrm{V}=\frac{5.8}{198}\mathrm{\hspace{0.17em}}\times \hspace{0.17em}\hspace{0.17em}6\hspace{0.17em}\hspace{0.17em}\times \hspace{0.17em}\hspace{0.17em}1000\mathrm{\hspace{0.17em}}\left(\mathrm{since} \mathrm{n}-\mathrm{factor}\hspace{0.17em}\hspace{0.17em}\mathrm{for}\hspace{0.17em}\hspace{0.17em}{\mathrm{NaIO}}_3=6\right)\\ =175.76\\ \frac{{\mathrm{W}}_{{\mathrm{NaHSO}}_3}}{{\mathrm{M}}_{{\mathrm{NaHSO}}_3}}\times \hspace{0.17em}\hspace{0.17em}2\hspace{0.17em}\hspace{0.17em}\times \hspace{0.17em}\hspace{0.17em}1000\mathrm{\hspace{0.17em}}=175.76\\ \therefore \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{W}}_{{\mathrm{NaHSO}}_3}\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}\frac{175.76\mathrm{\hspace{0.17em}}\times \hspace{0.17em}\hspace{0.17em}104}{2000}\mathrm{\hspace{0.17em}}=\mathrm{\hspace{0.17em}}9.14\hspace{0.17em}\hspace{0.17em}\mathrm{g}\end{array}$