Question

$MnO^{-}_4$ oxidises $(i)$ oxalate ion in acidic medium at $333\, K$ and $(ii) HCl$. For balanced chemical equations, the ratios $[MnO^{-}_4 : C_2O^{2-}_4]$ in $(i)$ and $[MnO^{-}_4 : HCl]$ 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)

$2MnO^{-}_4+ 5C_2O^{2-}_4 \overset{16 H^+}{\ce{->}}$
$2Mn^{2+}+ 10CO_2 + 8H_2O$
As $2$ moles of $MnO^{-}_4$ oxidises $5$ moles of $C_2O^{2-}_4$
$\therefore MnO^{-}_4 : C_2O^{2-}_4 = 2 : 5$
$2MnO^{-}_4 + 16HCl\rightarrow2MnCl_2 + 5Cl_2 + 2Cl^{-} + 8H_2O$
As $2$ moles of $MnO^{-}_4$ oxidises $16$ moles of $HCl$.
$\therefore MnO^{-}_4 : HCl = 2 : 16= 1: 8$