Question

If 33.1 g of lead nitrate is heated, calculate the mass of lead oxide and the volume of nitrogen and oxygen obtained at S.T.P.

Answer 1

Balanced Chemical Equation:

\[\ce{2Pb(NO3)2 -> 2PbO + 4NO2 + O2}\]

Molar Masses:

Pb(NO₃)₂ = 207 + 2[14 + (16 × 3)] = 331 g/mol

PbO = 207 + 16 = 223 g/mol

Moles of Pb(NO₃)₂, heated = `(33.1  g)/(331  g//mol)`

= 0.1 mole

1. Mass of Lead Oxide (PbO):

From the equation, 2 moles of Pb(NO₃)₂ produce 2 moles of PbO (1 : 1 ratio).

Moles of PbO formed = 0.1 mole

Mass of PbO = 0.1 × 223 = 22.3 g 

2. Volume of Nitrogen Dioxide (NO₂) at S.T.P.:

From the equation, 2 moles of Pb(NO₃)₂ produce 4 moles of NO₂.

Moles of NO₂ formed = 0.1 × 2 = 0.2 mole

Volume of NO₂ = 0.2 × 22.4 L = 4.48 L (or 4480 cm³)

3. Volume of Oxygen (O₂) at S.T.P.:

From the equation, 2 moles of Pb(NO₃)₂ produce 1 mole of O₂.

Moles of O₂ formed = `0.1/2` = 0.05 mole

Volume of O₂ = 0.05 × 22.4 L = 1.12 L (or 1120 cm³)