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.
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?
1 answer
1 answer