What is the stoichiometric coefficient of SO_2 in the following...

Oxidation half reaction: $\mathrm{SO}_2(\mathrm{ g})+2 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{HSO}_4^{-}(a q)+3 \mathrm{H}^{+}(a q)+2 e^{-}...(i)$ Reduction half reaction: $\mathrm{MnO}_4^{-}(a q)+8 \mathrm{H}^{+}+5 e^{-} \longrightarrow \mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_2 \mathrm{O}(l)...(ii)$ Multiply eq. (i) by 5 and (ii) by 2 and then add $$\begin{aligned} 2 \mathrm{MnO}_4^{-}(a q)+5 \mathrm{SO}_2(g) & +2 \mathrm{H}_2 \mathrm{O}(l)+\mathrm{H}^{+}(a q) \\ & \longrightarrow 2 \mathrm{Mn}^{+}(a q)+5 \mathrm{HSO}_4^{-}(a q) \end{aligned}$$ Thus, the stoichiometric coefficient of $\mathrm{SO}_2$ in the above balance equation is 5.