Question

I was doing a lab experiment on the reaction of copper(II) sulphate and aluminum:
1) I measured 2.02g of copper(II) sulphate pentahydrate
2) I dissolved the copper(II) sulphate pentahydrate in 10mL of distilled water
3) I added 2.0mL of concentrated HCl to the solution and mixed well
4) I added 0.25g on Aluminun foil to the solution.
5) After 5 minutes, i added an additional 5mL of concentrated HCl
6) After all the aluminum foil has reacted, i decanted the solution from the solid (leaving copper behind).
7) The mass of copper product after drying it was 0.69g.


Now we have:
-Mass of CuSO4*5H2O = 2.02g
-Mass of Aluminuim foil = 0.25g
-Mass of copper metal product = 0.69g
The questions are:

1) Calculate moles of Al used.
2) Calculate the moles of CuSO4*5H2O used.
3) Calculate the moles of copper product based on moles of Al.
4) Calculate the moles of copper product based on moles of CuSO4*5H2O.
5) Whats the limiting reactant.
6) Whats the grams of copper product based on the limiting reactant (theoretical yiel).
7) Calculate the percent yield of Copper.

Answer 1

And where are you on this problem; i.e., how much do you know how to do.
To get started, mols Al = grams/atomic mass = 0.25/26.98 = ?

Answer 2

0.00113

Answer 3

I need answer