A) 4/5
B) 3/5
C) 2/5
D) 1
Correct Answer: B
Solution :
When \[KMn{{O}_{4}}\] reacts with ferrous oxalate in acidic solution then the following reaction takes place. \[\overset{+7}{\mathop{Mn}}\,O_{4}^{-}+\overset{+2}{\mathop{Fe}}\,\overset{+3}{\mathop{{{C}_{2}}}}\,{{O}_{4}}\xrightarrow{{}}M{{n}^{+{{2}_{2+}}}}+2{{\overset{+4}{\mathop{CO}}\,}_{2}}+F{{e}^{3+}}\] Here, \[M{{n}^{7+}}\] is reduced to \[M{{n}^{2+}}(+7\to +2)\] and carbon is oxidised to \[C{{O}_{2}}\,(+3\to +4)\] and \[F{{e}^{3+}}\] is oxidised to \[F{{e}^{3+}}(+2\to +3)\]. Thus, Oxidation \[\left. \begin{matrix} C_{2}^{+3}\xrightarrow{{}}2\overset{+4}{\mathop{C}}\,{{O}_{2}}+2{{e}^{-}}\uparrow 2 \\ F{{e}^{2+}}\xrightarrow{{}}F{{e}^{3+}}+{{e}^{-}}\uparrow \,\,1 \\ \end{matrix} \right\}\uparrow \,3\] Reduction \[M{{n}^{7+}}+5{{e}^{-}}\xrightarrow{{}}M{{n}^{2+}}\downarrow 5\] The balanced chemical equation is as \[\underset{3\text{ }mol}{\mathop{\overset{+7}{\mathop{3MnO_{4}^{-}}}\,}}\,+\underset{\begin{smallmatrix} ferrous \\ oxalate \\ 5\text{ }mol \end{smallmatrix}}{\mathop{5Fe{{C}_{2}}{{O}_{4}}}}\,+24{{H}^{+}}\xrightarrow{{}}\overset{+2}{\mathop{3M{{n}^{2+}}}}\,\] \[+5\,F{{e}^{3+}}+10{{\overset{+4}{\mathop{CO}}\,}_{2}}+12{{H}_{2}}O\] \[\therefore \] 5 moles of \[Fe{{C}_{2}}{{O}_{4}}\] are oxidised by 3 moles of \[KMn{{O}_{4}}\]. \[\therefore \] 1 mole of \[Fe{{C}_{2}}{{O}_{4}}\] is oxidised by \[\frac{3}{5}\] moles of \[KMn{{O}_{4}}\].
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