# Aluminum and Copper(II) Chloride

## 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?

## 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?

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Borek
Mentor
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.

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

Borek
Mentor
Yes. Coefficients don't say anything about the amounts - only about their ratios during reaction.

Yes. Coefficients don't say anything about the amounts - only about their ratios during reaction.
Thanks for the help :)