2Pb (NO3)_{2} {\overset{ \triangle }{\longrightarrow}} 2PbO + 4NO_{2} + O_{2}

662 g of lead nitrate gives 89.6 lit. of NO_{2}

∴ 8.5 g of lead nitrate will give = 89.6\over 662 × 8.5

= 1.15 lit. of NO_{2}

Hence, **NO _{2} produced is 1.15 lit**

662 g of lead nitrate gives 22.4 lit. of O_{2}

∴ 8.5 g of lead nitrate will give 22.4\over 662 × 8.5

= 0.287 lit of O_{2}

Hence, **O _{2} produced is 0.287 lit**

∴ Total volume produced = 1.15 + 0.287 = 1.437 lits.

Hence, **total volume of NO _{2} and O_{2} produced on heating 8.5 of lead nitrate is 1.437 lit.**