Asked by jvjvukb
2.45 mol of NaOH are added to 45.0 g of copper (II) sulfate. How many moles of copper (II) hydroxide will precipitate out?
Answers
Answered by
DrBob222
2NaOH + CuSO4 ==> Cu(OH)2 + Na2SO4
mols NaOH = 2.45
mols CuSO4 = 45.0/159.5 = 0.282
2.45 mols NaOH with excess CuSO4 will produce 2.45/2 = 1.22 mols Cu(OH)2.
0.282 mols CuSO4 with excess NaOH will produce 0.282 mols Cu(OH)2.
This is a limiting reagent (LR) problem.In LR problems the smaller product wins since you can only get the smaller amount. So you will produce 0.282 mols of Cu(OH)2.
mols NaOH = 2.45
mols CuSO4 = 45.0/159.5 = 0.282
2.45 mols NaOH with excess CuSO4 will produce 2.45/2 = 1.22 mols Cu(OH)2.
0.282 mols CuSO4 with excess NaOH will produce 0.282 mols Cu(OH)2.
This is a limiting reagent (LR) problem.In LR problems the smaller product wins since you can only get the smaller amount. So you will produce 0.282 mols of Cu(OH)2.
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.