Using the Ion-Electron Method to balance the redox reaction:

1. Reduction Half-reaction: $Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$

• (Oxidation state of $Mn$ changes from $+7$ to $+2$, requiring $5e^{-}$).

2. Oxidation Half-reaction: $F{e}^{2+}\to F{e}^{3+}+e^{-}$

• (Oxidation state changes from $+2$ to $+3$, releasing $1e^{-}$).

3. Electron Balance: To equalize electrons, multiply the oxidation half-reaction by $5$.

• Result: $1Mn{O}_4^{-}$ reacts with $5F{e}^{2+}$.
• Therefore, the stoichiometric ratio is 1:5.