Question

How do we formulate the reaction between aluminum and chlorine gas? What is the molar equivalence?

Answer 1

We wants to make `AlCl_3`.........

Explanation:

Aluminum metal comes from Group 13 of the Periodic Table; it typically forms the `Al^(3+)` cation on oxidation......Meanwhile chlorine comes from Group 17. It is typically reduced to give the chloride ion, `Cl^-`.

And so `Al^(3+) + 3xxCl^-` and we gets `AlCl_3` as the neutral salt; in the solid phase it is probably the chloro-bridged dimer, `{Cl_2Al}_2(mu"-"Cl)_2`....