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

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Discussion Overview

The discussion revolves around calculating the theoretical yield of tetraaminecopper (II) sulfate hydrate from a reaction involving copper (II) sulfate pentahydrate, ammonia, water, and ethyl alcohol. Participants explore the stoichiometry of the reaction and the identification of limiting reagents.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant asks how to calculate the theoretical yield based on 1.00 g of CuSO4 and the amounts of other reactants used.
  • Another suggests converting all reactants into moles and identifying the limiting reagent to determine the yield.
  • Some participants propose that concentrated NH3 should be considered in excess, implying it does not limit the reaction.
  • There is uncertainty about whether ammonia is the limiting reagent, with differing opinions on the necessity of its consideration in calculations.
  • One participant speculates that CuSO4 might be the limiting agent but acknowledges the need for mole conversion to confirm this.
  • Another participant emphasizes that the limiting reagent can be determined by comparing the amounts of CuSO4 and NH3 used.

Areas of Agreement / Disagreement

Participants express differing views on whether concentrated ammonia should be treated as a limiting reagent or in excess. The discussion remains unresolved regarding the identification of the limiting reagent and the calculation of theoretical yield.

Contextual Notes

Participants note the importance of converting reactants to moles and the implications of using concentrated NH3, but there are unresolved assumptions regarding the definitions and calculations involved.

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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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
 
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.
 
ammonia is a limiting agent this time?
 
how many moles do you have of each starting material?
 
?
Was I wrong?
 
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.
 
That helps, buddy.
 
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...
 

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