Question

If $x$ moles of $Mn{O}_4^{-}$ required to oxidise $1$ mole of $Fe{\left(H{C}_2{O}_4\right)}_2$ in acidic medium and $y$ moles of $Ca{\left(OH\right)}_2$ are required to react with $1$ mole of $Fe{\left(H{C}_2{O}_4\right)}_2$, then determine the value of $x$ and $y$.

• A. $x=1,y=1$
• B. $x=1,y=2$
• C. $x=2,y=3$
• D. None of the above

Answer 1

The correct option is A $x=1,y=1$

$Fe{\left(H{C}_2{O}_4\right)}_2\rightleftharpoons {Fe}^{2+}+{2HC}_2{O}_4^{-}$

$Mn{O}_4^{-}$ oxidises ${Fe}^{2+}$ to ${Fe}^{2+}$ and ${HC}_2{O}_4^{-}$ to ${CO}_2$.

$Mn{O}_4^{-}+5{Fe}^{2+}⟶5{Fe}^{3+}+{Mn}^{2+}$
$2Mn{O}_4^{-}+5C_2{O}_4^{-}⟶10{CO}_2+2{Mn}^{2+}$

$5{Fe}^{2+}=1Mn{O}_4^{2-}$
$1{Fe}^{2+}=\frac{1}{5}Mn{O}_4^{2-}$
$5{HC}_2{O}_4^{-}=2Mn{O}_4^{-}$
$2{HC}_2{O}_4^{-}=\frac{4}{5}Mn{O}_4^{-}$

Total $Mn{O}_4^{-}$ required by $1$ mol of $Fe{\left(H{C}_2{O}_4\right)}_2$ $=\frac{1}{5}+\frac{4}{5}=1$ mol

$Fe{\left(H{C}_2{O}_4\right)}_2$ has two acidic $H^{+}$.

$Fe{\left(H{C}_2{O}_4\right)}_2\rightleftharpoons {Fe}^{2+}+2{HC}_2{O}_4^{-}\leftrightharpoons 2H^{+}+2C_2{O}_4^{-}$

$2H^{+}+Ca{\left(OH\right)}_2⟶{Ca}^{2+}+2H_{2}O$

Thus, $Ca{\left(OH\right)}_2$ required by $1$ mol of $Fe{\left(H{C}_2{O}_4\right)}_2=1$ mol