Question
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
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.