${\mathrm{KMnO}}_4$can be prepared from${\mathrm{K}}_2{\mathrm{MnO}}_4$as per the reaction

$3{\mathrm{MnO}}_4^{2-}+2{\mathrm{H}}_2\mathrm{O}\rightleftharpoons 2{\mathrm{MnO}}_4^{-}+{\mathrm{MnO}}_2+4{\mathrm{OH}}^{-}$

The reaction can go to completion by removing ${\mathrm{OH}}^{-}$ ions by adding

HCl and ${\mathrm{SO}}_2$ are reducing agents. So they can reduce  ${\mathrm{MnO}}_4^{-}. {\mathrm{CO}}_2$is neither oxidising nor reducing agent, it will provide only acidic medium. It can shift the reaction in forward direction and the reaction can go to completion.