Asked by Kiro
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.
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.
Answers
Answered by
DrBob222
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.
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.