Equivalent weight of  Pb(NO3)2 in the equation2Pb(NO3)2→∆ 2PbO+4NO2+O2 is

2Pb(N+5O-23)2 â†’∆ 2PbO+4N+4O2 +O02'N' atom Undergoes reduction'O' atom Undergoes OxidationTotal increase in Ox. state of 'N' per 2 moles of lead nitrate = 20-16=4EW of Pb(NO3)2 = 2M/4=M/2Alternate method:                 Total decrease in Ox. state of Oxygen per 2 moles of lead nitrate = [-20-(-24)]=4EW of Pb(NO3)2 = 2M/4=M/2