Question

CuSO4+2NsOH--> Cu(OH)2+Na2SO4

Answer 1

I got that the ratio of moles of CuSo4 to moles of NaOh is 1:4.
and that if 638.44g CuSO4 reacts with 240.0 NaOH, the limiting reagent is NaOH because CuSO4 is 4 moles and NaOH is 6 moles, but I need to use the limiting reagent to determine how many grams of Cu(OH)2 should precipitate out in the reaction

Answer 2

The reaction is this:
\[CuSO _{4} + 2NaOH \rightarrow Cu(OH)_{2} + Na _{2}SO _{4}\]
Therefore, one mole of copper(II) sulfate reacts with 2 moles of sodium hydroxide to form one mole of copper(II) hydroxide and one mole of sodium sulfate. You say that you are reacting 638.44g of copper(II) sulfate with 240.0g of sodium hydroxide. Using their molar masses, this gives us 5.0057 moles of copper(II) sulfate and 6.0006 moles of sodium hydroxide.

This means that, as you concluded, sodium hydroxide is the limiting reagent.

We can also see that from the reaction equation that 2 moles of sodium hydroxide react to form one mole of copper(II) hydroxide. This means that the number of moles of copper(II) hydroxide is half the number of moles of sodium hydroxide, or 3.0003 moles of copper(II) hydroxide.

Answer 3

5.0057 g of Cu(OH)2 is the precipitate of the reaction?

Answer 4

I get 292.709g of copper(II) hydroxide. There are 3.0003 moles of it, and it's molar mass is (15.999 + 1.008) * 2 + 63.546.

Answer 5

thanks so much!!

Answer 6

Sure! Glad I could help.