# A 3.10-g sample of lead nitrate, Pb(NO3)2, molar mass = 331 g/mol, is heated in an...

A 3.10-g sample of lead nitrate, Pb(NO3)2, molar mass = 331 g/mol, is heated in an evacuated cylinder with a volume of 2.00 L. The salt decomposes when heated, the unbalanced equation for the decomposition reaction is shown below: Pb(NO3)2(s) → PbO(s) + NO2(g) + O2(g) Assuming complete decomposition, what is the pressure in the cylinder after decomposition and cooling to a temperature of 303 K? Assume the PbO(s) takes up negligible volume.

Molar mass of Pb(NO3)2 = 331 g/mol

mass of Pb(NO3)2 = 3.1 g
mol of Pb(NO3)2 = (mass)/(molar mass)
= 3.1/331
= 9.37*10^-3 mol

Balanced chemical equation is:
2Pb(NO3)2 ---> 2 PbO + 4 NO2 + O2

According to balanced equation
mol of gas formed = (5/2)* moles of Pb(NO3)2
= (5/2)* 9.37*10^-3
= 2.34*10^-2 mol

Given:
V = 2.0 L
n = 0.0234 mol
T = 303.0 K

use:
P * V = n*R*T
P * 2 L = 0.0234 mol* 0.08206 atm.L/mol.K * 303 K
P = 0.2909 atm

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