2Pb (NO3)2 {\overset{ \triangle }{\longrightarrow}} 2PbO + 4NO2 + O2
662 g of lead nitrate gives 89.6 lit. of NO2
∴ 8.5 g of lead nitrate will give = 89.6\over 662 × 8.5
= 1.15 lit. of NO2
Hence, NO2 produced is 1.15 lit
662 g of lead nitrate gives 22.4 lit. of O2
∴ 8.5 g of lead nitrate will give 22.4\over 662 × 8.5
= 0.287 lit of O2
Hence, O2 produced is 0.287 lit
∴ Total volume produced = 1.15 + 0.287 = 1.437 lits.
Hence, total volume of NO2 and O2 produced on heating 8.5 of lead nitrate is 1.437 lit.