A3.31-g sample of lead nitrate, pb(no3)2, molar mass = 331 g/mol, is heated in an evacuated cylinder with a volume of 2.11 l. the salt decomposes when heated, according to the equation:

The equation in this problem is: 2Pb(NO3)2(s) --> 2PbO(s) + 4NO2(g) + O2(g)

And the question is what the pressure in the cylinder is after decomposition and cooling to a temperature of 300 K.

Solution:

Moles of Pb (NO3)2 = 3.31/331 = 0.0100 

2 moles of Pb (NO3)2 will decay to mold 4 moles of NO2 and 1 mole of O2. So 0.0100 moles of Pb (NO3)2 will form 0.02 moles of NO2 and 0.00500 moles of O2 

Then use the formula: PV = nRT. 

P = (0.02 + 0.005) * 0.082 * 300 / 1.62 

= 0.380 atm

the answer is barium and copper bromide