A) 3/5
B) 2/5
C) 4/5
D) 1
Correct Answer: A
Solution :
\[\underset{3M}{\mathop{3MnO_{4}^{-}}}\,+\underset{5M}{\mathop{5(F{{e}^{2+}}+{{C}_{2}}O_{4}^{2-})}}\,+24{{H}^{+}}\to \]\[3M{{n}^{2}}+5F{{e}^{3+}}+10C{{O}_{2}}+12{{H}_{2}}O\] Thus 5M of \[Fe{{C}_{2}}{{O}_{4}}\] is oxidised by 3M of \[KMn{{O}_{4}}\] then 1M of \[Fe{{C}_{2}}{{O}_{4}}\] is oxidised by 3/5 mole of \[KMn{{O}_{4}}\].
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