A 3.82-g sample of lead(II) nitrate, Pb(NO₃) ₂, molar mass = 331 g/mol, is heated in an evacuated cylinder with a volume of 1.70 L. The salt decomposes when heated, according to the equation
2Pb(NO₃) ₂(s) → 2PbO(s) + 4NO₂(g) + O₂(g)
Assuming complete decomposition, what is the pressure in the cylinder after decomposition and cooling to a temperature of 290. K? Assume the PbO(s) takes up negligible volume.

Question

Quizplus · Accepted Answer

The moles of Pb(NO3)2 are calculated as 3.82 g / 331 g/mol = 0.01154 mol. From the balanced equation, 2 moles of Pb(NO3)2 produce 5 moles of gas (4 NO2 + 1 O2). Therefore, 0.01154 mol of Pb(NO3)2 produces 0.01154 * (5/2) = 0.02885 mol of gas. Using the ideal gas law, PV = nRT, where R = 0.0821 L·atm/mol·K, T = 290 K, and V = 1.70 L, the pressure P = (0.02885 mol * 0.0821 L·atm/mol·K * 290 K) / 1.70 L = 0.404 atm.

A 382-G Sample of Lead(II) Nitrate, Pb(NO3)2, Molar Mass = 331

A 382-G Sample of Lead(II) Nitrate, Pb(NO₃)₂, Molar Mass = 331