Question

What weight of nitrate ion (calculated as $HNO_{3}$) is needed to convert 5 g of iodine into iodic acid according to the reaction
$I_{2} + HNO_{3} \to HIO_{3} + NO_{2} + H_{2}O$

• A. $12.205$ g
• B. $24.8$ g
• C. $0.248$ g
• D. $49.6$ g

Answer 1

Answer: (A)

From the given reaction, the balanced equation is
$I_{2}+ 10NO_{3}^{-} + 8H^{+} \to 2IO_{3}^{-}+ 10 NO_{2} + 4H_{2}O$
From the above equation,$254\, g$ of $I_2$ requires $620\, g$ of $NO_{3}^{-}$ ion
$\therefore 5 g of I_{2} $ requires $620$ of $NO_{3}^{-}$ ion$=\frac{620\times5}{254}$
$=12.205$ g of $NO_{3}^{- }$