Calculating CuSO4*5H2O Yield from H2SO4 and Copper Reaction

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Discussion Overview

The discussion revolves around calculating the yield of CuSO4*5H2O from the reaction between sulfuric acid (H2SO4) and copper (Cu). Participants explore the stoichiometry of the reaction, the nature of the chemical processes involved, and the correct representation of the reaction, including redox considerations.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant calculates the yield of CuSO4*5H2O based on the amount of H2SO4 used and the efficiency of crystallization, arriving at a value of 42.39 g.
  • Another participant challenges the reaction's representation, stating it is a redox reaction and suggesting the initial formulation is incorrect.
  • Further contributions attempt to clarify the oxidation and reduction processes, with one participant noting that copper is oxidized and sulfuric acid acts as the oxidizer.
  • There are multiple incorrect formulations of the reaction presented, indicating confusion about the products and stoichiometry involved.
  • One participant expresses uncertainty about identifying the oxidizer in the reaction.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct representation of the reaction or the stoichiometric calculations. There are competing views on the nature of the reaction and its products.

Contextual Notes

Participants express uncertainty regarding the correct redox processes and the identification of reactants and products. There are unresolved issues related to the stoichiometry and the nature of the chemical reaction.

danne89
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Hi again! The problem goes like this:
How many g CuSO4*5H_2O does one get when 20 g of a 98% H2SO4 solution and copper interact, if only 85% of the cristalls is possible to use.

[tex]Cu+H_2SO_4 + 5H_2O \rightarrow CuSO_4 * 5H_2O + H_2[/tex]
20*0.98=19.6
[tex]\frac{19.6}{H_2SO_4} = \frac{x}{CuSO_4*5H_2O}[/tex]
x=49.87
49.87 * 0.85 = 42.39
 
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I'm sorry,but the reaction is REDOX AND IT IS NOT CORRECT IN THE FORM U HAVE GIVEN.Please adjust it.

Daniel.
 
I'm sorry too; for being so stupid.
[tex]Cu \rightarrow Cu^{2+} + 2e^-[/tex]
[tex]H_2SO_4 + 2e^- \rightarrow H_2 + SO_4^{2-}[/tex]
[tex]SO_4^{2-} + Cu^{2+} \rightarrow CuSO_4[/tex]
[tex]CuSO_4 + 5H_2O \rightarrow CuSO_4 + 5 H_2O[/tex]
Hmm. That cannot be right. Cu is a preciouser metal then H...
 
That's right.From your reaction,there will not come out any Hydrogen.Take another chance.

Daniel.
 
[tex]Cu + 5 H_2SO_4 \rightarrow CuSO_4 * 5H_2O +5 SO_4^{2-}[/tex]

Edit: No! It's wrong too. Bluaa
 
Last edited:
No.It's wrong.If i tell u the reaction products,can i trust you'll do the redox thing...?

Daniel.

P.S.Copper(II) sulphate,sulphur dioxyde and plain water.
 
Nooo. I can't. I even cannot locate the oxidizer.
 
Sulphate ion is the oxydizer (the sulphur gets reduced).Okay,i'll right the reaction and u'll do the REDOX THING...:wink:

[tex]Cu+2H_{2}SO_{4}\rightarrow CuSO_{4}+SO_{2}\uparrow +2H_O[/tex]

You can solve your problem now,but it would be nice if you figured out how this reaction is trully going...

Daniel.
 

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