So if it's relatively pure copper sulfate, then absolutely use iron to cement the copper. Just make sure the solution is dilute enough that the copper won't form an impermeable crust around the iron.
You can also do what I do for my larger volume of waste, and use a simple shorted galvanic cell setup: a big iron rod on one side of the solution, a copper rod on the other, and a copper wire connecting the two held outside the solution. This sets up an internal electric current that rapidly reduces copper from solution onto both the iron AND the copper rods.
The solution does need to be at an acid pH for it to work. Right around pH 2 seems to work very well. I usually acidify with HCl, as the presence of a little chloride ion appears to facilitate the process. Sulfate solution alone is rather slower, probably due to complicated electrochemical ion potential stuff I can't explain, cuz it's been 25 years since chemistry class, lol.
Thanks and appreciate your response, I'd still drop the CU with FE powder and make ferrous sulfate as a sellable, which I'll look into should I find the right buyer. Right now, I'm kind of open to any sulfate(Zinc, Aluminum, Iron) as long as it stays on the economical side.
The OP is trying to displace the Copper with Aluminum and thus have two salebale products.
Copper and Aluminum Sulfate.
I do not know if this is feasible economically though.
Thanks for clearing it out once again
Although I'm completely hopeful, fingers still crossed
On the other hand I am done with my first batch of
CuSo45h20 125g
Aluminum consumed 11.59g
Yield 32g
Great on this end
However, to understand the stoichiometry better (as I got 13.58g of AL required to drop the aforementioned copper), I did a little search again and look what I found, three different responses by different chemists for the same question:
How many grams of solid copper will theoretically be produced when 2.2G of aluminum is reacted with 14.4G of copper (II) sulfate? Which reactant is the limiting reactant? Show work. Be sure to include units. 2Al+3CuSo4→3Cu+Al2(SO4) 3
1,
Al=27. . . . . CuSO4= 160……. . . . .Cu:64
2Al. + 3CuSo4–––>3Cu+Al2(SO4) 3
54g………480g. .. ..…192g
2.2g……….14.4g Limiting reactant is CuSO4 because :
2.2/54 ratio is higher than 14.4/480 ratio. Al is excess.
Copper to be produced:
480 g CuSO4 –––––> 192 g of Cu
14.4 g CuSO4––––––> x
x = 14.4*192 / 480 =
5.76 g of Cu
2,
find moles of substances using the equation: n=m/M so for copper sulfate it would be n=14.4/159.6= 0.09 mols
then since the ratio of copper sulfate to copper is 3:3, the moles would be the same (0.09 mol)
this time rearrange the formula into m=nM so m=0.09x342.15=
30.8 grams
3,
2 moles of Al = 2×27g =54g
3 moles of CuSO4 = 3(63.5+32+4×16)g= 478.5g
3 moles of Cu = 3×63.5g = 190.5g
54g of Al reacts with 478.5g of CuSO4
So,2.2g………………………478.5×2.2/54= 19.49g of CuSO4
But only 14.4g of copper(ii) sulphate added in the reaction vessel. Thus,CuSO4 is the limiting reactant.
Now,54g of Al produces 190.5g of Cu
Therefore,2.2g………………..190.5×2.2/54=
7.76g of Cu
How do I even understand which one of them are correct to double check my mistake???
I really need help understanding the calculation regarding this reaction.
Many thanks in advance