Correct option is (A) 6/5
In acidic medium-
\(M_n{O_4}^- \longrightarrow Mn^{+2}\)
or
\(Mn{O_4}^- + 8H^+ + 5e^- \longrightarrow Mn^{+2} + 4H_2O\) ....(i)
\([C_2{O_4}^{-2} \longrightarrow 2CO_2 + 2e^-]\) ......(ii)
Applying 5 x (ii) + 2 x (i), we get
⇒ \(2Mn{O_4}^- + 16H^+ + 5C_2 {O_4}^{-2} \longrightarrow 10CO_2 + 2Mn^{+2} + 8H_2O\)
or
\(2KMnO_4 + 8H_2SO_4+ \frac 53 Fe_2(C_2O_4)_3 \longrightarrow 10CO_2 + 2MnSO_4 + K_2SO_4 + \frac 53Fe_2(SO_4)_3\)
\(\because \frac 53\) mole Fe2(C2O4)3 react with 2 mole of KMnO4.
\(\therefore\) 1 mole of Fe2(C2O4)3 react with = \(\frac 2{5/3}\) mole of KMnO4
\(= \frac{2\times 3}5\)
\(= \frac 65\) mole of KMnO4
