Step 1: {Oxidation Half Reaction}
Write the oxidation half reaction:
\[
{SO}_2(g) + 2 {H}_2{O}(l) \rightarrow {HSO}_4^-(aq) + 3 {H}^+(aq) + 2e^-
\]
Step 2: {Reduction Half Reaction}
Write the reduction half reaction:
\[
{MnO}_4^-(aq) + 8 {H}^+(aq) + 5e^- \rightarrow {Mn}^{2+}(aq) + 4 {H}_2{O}(l)
\]
Step 3: {Balance the Electrons}
To balance the electrons, multiply the oxidation half reaction by 5 and the reduction half reaction by 2:
\[
5 {SO}_2(g) + 10 {H}_2{O}(l) \rightarrow 5 {HSO}_4^-(aq) + 15 {H}^+(aq) + 10e^-
\]
\[
2 {MnO}_4^-(aq) + 16 {H}^+(aq) + 10e^- \rightarrow 2 {Mn}^{2+}(aq) + 8 {H}_2{O}(l)
\]
Step 4: {Combine and Simplify}
Combine the two half reactions:
\[
2 {MnO}_4^-(aq) + 5 {SO}_2(g) + 16 {H}^+(aq) \rightarrow 2 {Mn}^{2+}(aq) + 5 {HSO}_4^-(aq) + 8 {H}_2{O}(l)
\]
Thus, the stoichiometric coefficient of SO$_2$ is 5.