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An other stoiciometry problem

  1. Feb 15, 2005 #1
    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
     
  2. jcsd
  3. Feb 15, 2005 #2

    dextercioby

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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.
     
  4. Feb 15, 2005 #3
    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...
     
  5. Feb 15, 2005 #4

    dextercioby

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    That's right.From your reaction,there will not come out any Hydrogen.Take another chance.

    Daniel.
     
  6. Feb 15, 2005 #5
    [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: Feb 15, 2005
  7. Feb 15, 2005 #6

    dextercioby

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    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.
     
  8. Feb 16, 2005 #7
    Nooo. I can't. I even cannot locate the oxidizer.
     
  9. Feb 16, 2005 #8

    dextercioby

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    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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