Answer :
Answer:
0.582 moles of aluminum chloride are produced
Explanation:
We determine the reaction to begin the excersise:
2Al (s) + 3Cl₂(g) → 2AlCl₃
As the chlorine is in excess, Al must be the limiting reactant. And the ratio is 2:2, so the moles I used from Al will be the same moles of chloride produced.
We convert the mass from Al to moles:
23 g . 1 mol / 26.98g = 0.852 moles
In conclussion we produce 0.582 moles of aluminum chloride