What weight of nitrate ion (calculated as HNO₃) is needed to convert 5 g of iodine into iodic acid according to the reaction

${\mathrm{I}}_2+{\mathrm{HNO}}_3⟶{\mathrm{HIO}}_3+{\mathrm{NO}}_2+{\mathrm{H}}_2\mathrm{O}$

From the given reaction, the balanced equation is -

${\mathrm{I}}_2+10{\mathrm{NO}}_3^{-}+8{\mathrm{H}}^{+}\to 2{\mathrm{IO}}_3^{-}+10{\mathrm{NO}}_2+4{\mathrm{H}}_2\mathrm{O}$

From the above reaction,

$254\mathrm{g}\text{ of }{\mathrm{I}}_2\text{ requires }620\mathrm{g}\text{ of }{\mathrm{NO}}_3^{-}\text{ion }$

$\begin{array}{l}\therefore 5\mathrm{g}\text{ of }{\mathrm{I}}_2\text{ requires }620\text{ of }{\mathrm{NO}}_3^{-}\text{ion }=\frac{620\times 5\mathrm{g}}{254}=12.205\mathrm{g}\text{ of }\\ {\mathrm{NO}}_3^{-}\end{array}$