Asked by Jefferson
Lead(II) carbonate decomposes to form Lead(II) oxide and carbon dioxide. If 5.oo grams of the reactant decomposes to produce 1.52 grams of Lead(II) oxide experimentally, what is the theoretical yield and percent yield.
Answers
Answered by
Devron
PbCO3 ---> PbO + CO2
First, calculate theoretical yield:
5.00g*(1 mole/267.21 g) = moles of PbCO3
moles of PbCO3=moles of PbO
moles of PbO*(223.20 g/mole)= mass of PbO
mass of PbO=theoretical yield
(1.52g/theoretical yield)*100= % Yield
First, calculate theoretical yield:
5.00g*(1 mole/267.21 g) = moles of PbCO3
moles of PbCO3=moles of PbO
moles of PbO*(223.20 g/mole)= mass of PbO
mass of PbO=theoretical yield
(1.52g/theoretical yield)*100= % Yield
Answered by
Anonymous
when heated lead nitrate decomposes to lead oxide, nitrogen dioxide and oxygen gas
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.