Question

metal halides. For example, 2 Al (s) + 3 Cl2 (g) ⟶ 2AlCl3 (s)

If you begin with 99.5g of aluminum

Answer 1

Moles of Al =$\frac{99.5}{26.98}=3.68 moles$

Moles of $Cl_2=\frac{3}{2}×3.68=5.52moles$

Moles of $AlCl_3= 3.68moles$