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

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In summary, the theoretical yield of product tetraaminecopper (II) sulfate hydrate from 1.00 g of copper (II) sulfate pentahydrate (CuSO4) is 3 moles.
  • #1
primarygun
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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!
 
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  • #2
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
 
  • #3
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.
 
  • #4
ammonia is a limiting agent this time?
 
  • #5
how many moles do you have of each starting material?
 
  • #6
?
Was I wrong?
 
  • #7
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.
 
  • #8
That helps, buddy.
 
  • #9
i can't tell if you are being sarcastic or not. if so, I am not doing your work for you. do it yourself
 
  • #10
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.
 
  • #11
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...
 

1. What is the purpose of this reaction?

The purpose of this reaction is to produce a yield of 1.00g of copper(II) sulfate (CuSO4) in the form of a tetrahydrate, using ammonia (NH3), water (H2O), and ethanol (C2H6O) as reactants.

2. What are the reactants in this reaction?

The reactants in this reaction are 1.00g of copper(II) sulfate, an unknown amount of ammonia, an unknown amount of water, and an unknown amount of ethanol.

3. What is the chemical equation for this reaction?

The chemical equation for this reaction is: CuSO4 + 4NH3 + 4H2O + 2C2H6O → CuSO4·4H2O + 4NH4+ + 2C2H5OH

4. What is the theoretical yield of copper(II) sulfate in this reaction?

The theoretical yield of copper(II) sulfate in this reaction is 1.00g.

5. What are the conditions necessary for this reaction to occur?

This reaction requires the presence of copper(II) sulfate, ammonia, water, and ethanol, as well as a suitable environment for the reaction to take place (e.g. appropriate temperature and pressure).

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