Calc Tetra Yield:1.00g CuSO4 + NH3 + H2O + Ethanol

[primarygun]

for the synthesis to be performed in this experiment, based on 1.00 g of copper (II) sulfate pentahydrate (CuSO4) taken initially, what is the theoretical yield of product tetraaminecopper (II) sulfate hydrate? [Cu(NH3)4]SO4 - H2O

the amounts of reactants I'm using in this experiment are:
1 g of CuSO4
10 mL of distilled water
5 mL of concentrated NH3 (what does concentrated mean? 1.0 M I'm assuming?)
and approximately 10 mL of ethyl alcohol

what i have set up as the equation so far:
CuSO4 + H2O + 4NH3 --> [Cu(NH3)4]SO4 - H2O

does anyone have any idea on how i can calculate theoretical yield of the product? any help/suggestion would be greatly appreciated!

 

[gravenewworld]

1) get everything into moles
2) find out what is your limiting reagent.
3) from stoichiometry figure out how much product your limiting reagent can produce if all of it reacts

 

[gravenewworld]

I'm pretty sure here, without doing any calculations, that concentrated NH3 means just ignore NH3 when trying to do this problem. just consider it as an reagent with an excess amount.

 

[primarygun]

ammonia is a limiting agent this time?

 

[gravenewworld]

how many moles do you have of each starting material?

 

[primarygun]

?
Was I wrong?

 

[gravenewworld]

yes. just ignore the ammonia. concentrated ammonia here means you have an excess amount of it. find out which reagent is going to get eaten up by the reaction first.

 

[primarygun]

That helps, buddy.

 

[gravenewworld]

i can't tell if you are being sarcastic or not. if so, I am not doing your work for you. do it yourself

 

[primarygun]

i can't tell if you are being sarcastic or not. if so, I am not doing your work for you. do it yourself

This was a question from other registers.
I gave suggestions to him but the question seems to be a bit faulty.
Therefore, I want to make sure I didn't give him a wrong direction.

 

[professor]

well here's a guess... the limiting agent is probably CuSO4 but i can't be sure unless i converted it to moles... basically, whatever you have less of, ammonia or CuSO4 is your limiting agent, because an equal amount of each is being used up to create a product molecule. then work from there once you know... this will give you a number of molecules equal to the limiting agent in this case it looks like (which will be about three times as many moles of substance) aka 1 mol of limiting agent = 3 mols of the terramine copper sulfur hydrate...correct me if anyone thinks otherwise...