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?