Aluminum and Copper(II) Chloride

  • Thread starter Jimbo57
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  • #1
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Homework Statement


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?

Homework Equations





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?
 

Answers and Replies

  • #2
Borek
Mentor
28,600
3,081
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
Where is it stated?

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?
Much better.
 
  • #3
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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
 
  • #4
Borek
Mentor
28,600
3,081
Yes. Coefficients don't say anything about the amounts - only about their ratios during reaction.
 
  • #5
96
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Yes. Coefficients don't say anything about the amounts - only about their ratios during reaction.
Thanks for the help :)
 

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