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Aluminum and Copper(II) Chloride

  1. May 21, 2012 #1
    1. The problem statement, all variables and given/known data
    3CuCl2(aq) + 2Al(s) --> 2AlCl3(aq) + 3Cu(s)

    a) If 4.8 mol of Al react in the process, how many moles of Cu are produced?
    b) How many grams of Al will be necessary in order to produce 12.7g of Copper?
    2. Relevant equations



    3. The attempt at a solution
    a) This one is tricky in that I originally did 4.8/2Al X 3Cu = 7.2mol of Cu but then I realized that there is still only 3mol of Cu present as a reactant regardless of the concentration of Al. So I would assume the 3Cu would stay the same.

    b) 12.7/63.5 gr Cu= 0.2mol Cu
    0.2mol Cu x 2Al/3Cu = 0.13mol Al
    0.13mol Al x 27gr/mol = 3.6gr Al for 12.7gr of Cu

    How does that look?
     
  2. jcsd
  3. May 21, 2012 #2

    Borek

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    Staff: Mentor

    Where is it stated?

    Much better.
     
  4. May 21, 2012 #3
    Hey Borek,

    I was assuming that 3CuCl2(aq) meant there was 3 mol copper present, but I'm guessing that isn't the case (I'm very new to this). So my original assumption is the better one?
    4.8/2Al X 3Cu = 7.2mol of Cu
     
  5. May 21, 2012 #4

    Borek

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    Staff: Mentor

    Yes. Coefficients don't say anything about the amounts - only about their ratios during reaction.
     
  6. May 21, 2012 #5
    Thanks for the help :)
     
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