Why do the following reactions proceed differently? $Pb_{3}O_{4} + 8HCl \rightarrow 3PbCl_{2} + Cl_{2} + 4H_{2}O$ and $Pb_{3}O_{4} + 4HNO_{3} \rightarrow 2Pb(NO_{3})_{2} + PbO_{2} + 2H_{2}O$

(N/A) $Pb_{3}O_{4}$ is a stoichiometric mixture of $2 \ mol$ of $PbO$ and $1 \ mol$ of $PbO_{2}$. In $PbO_{2}$,lead is in the $+4$ oxidation state,while in $PbO$,it is in the $+2$ oxidation state.
$PbO_{2}$ acts as an oxidizing agent and can oxidize the $Cl^{-}$ ion of $HCl$ to $Cl_{2}$. The reaction with $HCl$ is a combination of an acid-base reaction $(2PbO + 4HCl \rightarrow 2PbCl_{2} + 2H_{2}O)$ and a redox reaction $(PbO_{2} + 4HCl \rightarrow PbCl_{2} + Cl_{2} + 2H_{2}O)$.
In contrast,$HNO_{3}$ is itself an oxidizing agent,so it does not react with $PbO_{2}$. The reaction with $HNO_{3}$ is limited to the acid-base reaction between $PbO$ and $HNO_{3}$: $2PbO + 4HNO_{3} \rightarrow 2Pb(NO_{3})_{2} + 2H_{2}O$. The $PbO_{2}$ remains as a residue,making the overall reaction different.