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Chemistry

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]

Stoichiometry

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Answer

2Pb[NO3]2→ Δ 2PbO+4NO2+O22[207+2(14+48)](4×22.4)22.4 lit=2[207+124]=89.6 lit=662 g\begin{matrix} 2\text{Pb[NO}3]2 & \xrightarrow{\space \Delta \space} & 2\text{PbO} & + & 4\text{NO}2 & + & \text{O}2 \ 2[207 + 2(14 + 48)] & & & & (4 \times 22.4) & & 22.4 \text{ lit} \ = 2[207 + 124] & & & & = 89.6 \text{ lit} & & \ = 662 \text{ g} & & & & & & \end{matrix}

662 g of lead nitrate gives 89.6 lit. of NO2

∴ 8.5 g of lead nitrate will give 89.6662\dfrac{89.6}{662} x 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 22.4662\dfrac{22.4}{662} x 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 lits.

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