The experiment began with aluminum metal and copper (II) chloride solution. What substances were present in the beaker at the end of the experiment?
You have Aluminum and chlorine left. I think you want to know how they combine and the stoichiometry. Aluminum is Al. Coppoer chloride (ll) is CuCl2. 2 Al + 3 CuCl2 - 2 AlCl3 + 3 Cu Good luck.
Aluminum metal is notoriously nonreactive due to a passivating layer of Al2O3. In the presence of chloride ions, the layer dissolves, producing AlCl4^-, and Al metal, which reacts with Cu2+ ions. Cu metal is produced. You may find Cu metal, and Al metal, plus Cu2+, Cl-, Al3+, AlCl4^- ions. The amounts at the end depend on the amounts you start with. Al2O3(s) + 8Cl- + 3HOH -- 2AlCl4^- + 6OH- 2Al(s) + 3Cu2+ -- 2Al3+ + 3Cu(s) ..... Cl- is a spectator ion
