Question

$1$ g of a moist sample of a mixture of $KCl{O}_3$ and $KCl$ was dissolved in water and made upto $250$ mL. $25$ mL of this solution was treated with $S{O}_2$ to reduce chlorate into chloride and the excess $S{O}_2$ was boiled off. When the total chloride was precipitated, $0.1435$ g of AgCI was obtained. In another experiment, $25$ mL of the original solution was treated with $30$ mL of $0.2$ N solution of $FeS{O}_4$ and unreacted $FeS{O}_4$ required $37.5$ mL of $0.08$N solution of an oxidizing agent for complete oxidation. The molar ratio of chlorate and chloride in the given mixture $F{e}^{2+}$ reacts with $CI{O}_3^{-}$ according to equation, $Cl{O}_3^{-}+6F{e}^{2+}+6H^{+}\to C{I}^{-}+6F{e}^{3+}+3H_{2}O$, is :

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is A 1

$6H^{+}+Cl{O}_3^{-}+6e^{-}\to C{l}^{-}+3H_{2}O$
$6(F{e}^{2+}\to F{e}^{3+}+1e^{-})$
------------------------------------------------------------
$6H^{+}+Cl{O}_3^{-}+6F{e}^{2+}\to C{l}^{-}+6F{e}^{3+}+3H_{2}O$

Now as here, the ratio of moles of $Cl{O}_3^{-}$ on left side of the reaction and $C{l}^{-}$ is 1:1.