Question

`2Al + 3Cl_2 -> 2AlCl_3` How many grams of chlorine are required to produce 4 moles of aluminum chloride?

Answer 1

Approx. `425*g`

Explanation:

`2Al(s) + 3Cl_2(g) rarr 2AlCl_3(s)`

You have given a stoichiometrically balanced equation, so bravo.

The equation explicitly tells us that `54*g` of aluminum metal reacts with `6xx35.45*g` `Cl_2` gas to give `266.7*g` of `"aluminum trichloride"`. From where am I getting these numbers?