You didn't write the decomposition equation nor did you not if was Pb(II) nitrate or Pb(IV) nitratem. Is this it?
2Pb(NO3)2→2PbO+4NO2+O2
mols Pb(NO3)2 = grams/molar mass = 3.31/331 = 0.01
mols NO2 formed = 4 x 0.01 = 0.04
mols O2 formed = 1 x 0.01 = 0.01
Total mols gas = 0.05
Then use PV = nRT and solve for total pressure in the container. Post your work if you get stuck.
A 3.31-g sample of lead nitrate, , molar mass = 331 g/mol, is heated in an
evacuated cylinder with a volume of 2.37 L. The salt decomposes when heated, according to
the equation:
Assuming complete decomposition, what is the pressure in the cylinder after decomposition
and cooling to a temperature of 300. K? Assume the takes up negligible volume.
1 answer
1 answer