2.68 × 10 ^ - 3 moles of a solution containing an ion A^ n+ require...

\left[{\text{A}}^{\text{n} +} \to \overset{+ 5}{\text{A}} {\text{O}_{3} {}^{-} + \left(5 - \text{n}\right){\text{e}}^{-}\right] \times 5,$Oxidation half reaction<br /><br />\left[{MnO}_{4}^{-} + 8{H}^{+} + 5{e}^{-} \to {Mn}^{2 +} + 4{H}_{2}O\right] \times ( 5 - n ) ,$Reduction half reaction

Applying reaction stoichiometry,

{\text{5 moles of A}}^{\text{n+}} = \left(\text{5-n}\right){\text{moles of MnO}_{4}^{-}
$\dfrac{\text{5}}{\text{5-n}} = \dfrac{\text{2.68} \times 1 {0}^{- 3}}{\text{1.61} \times {\text{10}}^{- 3}} = \text{1.66} = \dfrac{\text{5}}{\text{3}}$
$5 - \text{n} = 3$
$∴ \text{n} = 2$