[stockyrock]

First Case:
Hello Everyone, can someone help me with that problem?

Cu2O + 2HCl --> 2CuCl + H20
3CuCl + Al --> AlCl3 + 3Cu

Second Case
CuO + 2HCl -- > CuCl2 + H2O
3CuCl2 + 2Al --> 2 AlCl3 + 3Cu

Add Infos:
This is a lab experiment,
we will use about 0.1g of Al and between .15g and .20g of Copper Oxide.
The purpose of the experiement is to identifiy the Copper oxide that our team got.

[Hunter2]

This experiment is nonsense, because Cu-I-compounds not very good soluble. In first case you will have a precipitate still and the conversion from Oxide to chloride will not be efficient. In second case you will get a good soluble copper chloride from blue-green color. From this observation you can directly tell what you got. The second part with the aluminum is not necessary.

Let assume you would have CuCl in solution by usinf complexagent you can see on your equations how much copper you can obtain with 1 mol aluminium.