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

We wants to make $A l C {l}_{3}$.........
Aluminum metal comes from Group 13 of the Periodic Table; it typically forms the $A {l}^{3 +}$ cation on oxidation......Meanwhile chlorine comes from Group 17. It is typically reduced to give the chloride ion, $C {l}^{-}$.
And so $A {l}^{3 +} + 3 \times C {l}^{-}$ and we gets $A l C {l}_{3}$ as the neutral salt; in the solid phase it is probably the chloro-bridged dimer, ${\left\{C {l}_{2} A l\right\}}_{2} {\left(\mu \text{-} C l\right)}_{2}$....