Question

$Mn{O}_4^{-}$ oxidises $(i)$ oxalate ion in acidic medium at $333K$ and $(ii)HCl$. For balanced chemical equations, the ratios $[Mn{O}_4^{-}:C_2{O}_4^{2-}]$ in $(i)$ and $[Mn{O}_4^{-}:HCl]$ in $(ii)$ respectively are

• A. 1 : 5 and 2 : 5
• B. 2 : 5 and 1 : 8
• C. 2 : 5 and 1 : 5
• D. 5 : 2 and 1 : 8

Answer 1

Answer: (B)

$2Mn{O}_4^{-}+5C_2{O}_4^{2-}\stackrel{16H^{+}}{\to }$
$2M{n}^{2+}+10C{O}_2+8H_{2}O$
As $2$ moles of $Mn{O}_4^{-}$ oxidises $5$ moles of $C_2{O}_4^{2-}$
$\therefore Mn{O}_4^{-}:C_2{O}_4^{2-}=2:5$
$2Mn{O}_4^{-}+16HCl\to 2MnC{l}_2+5C{l}_2+2C{l}^{-}+8H_{2}O$
As $2$ moles of $Mn{O}_4^{-}$ oxidises $16$ moles of $HCl$.
$\therefore Mn{O}_4^{-}:HCl=2:16=1:8$