Heat on lead nitrate gives yellow lead [II] oxide, nitrogen dioxide and oxygen. Calculate the total volume of NO2 and O2 produced on heating 8.5 of lead nitrate. [Pb = 207, N = 14, O = 16]

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.