oxidation - reduction doubt

Do it in this way...Pb₃O₄ is a mixed oxide of 2PbO and PbO₂....so Pb is changing it's oxidation state from +4 to +2 in PbO₂ and in PbO it's oxidation state remains same....

MnO₂ + PbO₂ +PbO + HNO₃ → Pb(NO₃)₂ + HMnO₄ + H₂O

so Pb is changing it's oxidation state from +4 to +2 and Mn from +4 to +7..so write the equation as .
so mutiply MnO₂ by 2 and PbO₂ by 3..
Now to form a mixed oxide the number or moles of PbO₂ = 2 * the number of moles of PbO..so multiply PbO by 6...

So u will get the balanced equation as
2MnO₂ + 3Pb₃O₄ + 18HNO₃ → 9Pb(NO₃)₂ + 2HMnO₄ + 8H₂O

THANKS GOVIND [1]