Why following two reaction proceed differently?
`Pb_(3)O_(4)+8HClrarr3PbCl_(2)+Cl_(2)+4H_(2)O`
and
`Pb_(3)O_(4)+4HNO_(3)rarr2Pb(NO_(3))_(2)+PbO_(2)+2H_(2)O`

Explanation

`Pb_(3)O_(4)` is actually a stoichiometric mixture of 2 moles of PbO and one mole of `PbO_(2). PbO_(2)` ion `PbO_(2)` lead is present in +4 oxidation state whereas in PbO lead is present in +2 oxidation state since oxdise `CI^(-)` ions of HCI in to `CI_(2)` furthermore PbO is a basic oxide and hecne reacts with HCI to form `PbCI_(2)` and `H_(2)O` thus the reactoin of `Pb_(3)O_(4)` with HCI can be split in to two reactions neamely acid base and redox reaction as shown below
`2 PbO + 4 HCI rarr 2PbCI_(2)+2H_(0)O`
`PbO_(2)+4HCIrarrHCIrarr3pbCI_(2)+CI_(2)+2H_(2)O`
`2PbO+PbO_(2)+8HCIrarr3PbCI_(2)+CI_(2)+4H_(2)O`
`Pb_(3)O_(4)+8 HCIrarr3PbCI_(2)+CI_(2)+4H_(2)O`
`Pb_(3)O_(4)+8HCIrarr3PbCI_(2)+CI_(2)+4H_(2)O`
sice `HNO_(3)` is an oxidising agent and in `PbO_(2),Pb` is ready present in its highest oxidation state of + 4 therefore no redox reaction occurs between `PbO_(2)` and `HNO_(3)` howver acid base reaction between pbO and `HNO_(3)` occurs as follows
`2 PbO + PbO_(2)+4HnO_(3)rarr2Pb(NO_(3))_(2)+PbO_(2)+2H_(2)O M`
`Pb_(3)O_(4)+4HNO_(3)rarr2Pb(NO_(3)^(2)+PbO_(2)+2H_(2)O`
thus it is the passive nature of `PbO_(2)` against `HNO_(3)` that makes the reaction of `Pb_(3)O_(4)` with `HNO_(3)` different from that with HCI