Question

$10^{-2}\, mol$ of $Fe _{3} O _{4}$ is treated with excess of $KI$ solution in the presence of dilute $H _{2} SO _{4}$, the products are $Fe ^{2+}$ and $I _{2}( g )$. What volume of $0.1( M ) Na _{2} S _{2} O _{3}$ will be needed to reduce the liberated $I _{2}( g )$ ?

• A. $50\, ml$
• B. $100\, ml$
• C. $200\, ml$
• D. $400\, ml$

Answer 1

Answer: (C)

$\therefore Fe _{3} O _{4}+2 I^- \rightarrow Fe ^{2+}+ I _{2}$
$\therefore $ No. of moles of $I _{2}$ produced $=10^{-2} \, mol$
Let $v \,ml \, 0.1( M ) Na _{2} S _{2} O _{3}$ solution is required
$\therefore v \times 10^{-4}=2 \times 10^{-2}$
or $ v=200 \,ml$.