Aluminium chloride can be prepared by passing dry chlorine or hydrogen chloride gas over heated aluminium metal. The equation for the chlorine reaction is:
2Al(s) + 3Cl2(g) == 2AlCl3(s)
How much aluminium is needed to make 89 g of aluminium chloride?
(Relative atomic masses: Al = 27.0, Cl = 35.5, H =1) [rmc-29]

Question

Aluminium chloride can be prepared by passing dry chlorine or hydrogen chloride gas over heated aluminium metal. The equation for the chlorine reaction is:
2Al(s) + 3Cl2(g) ==> 2AlCl3(s)
How much aluminium is needed to make 89 g of aluminium chloride?
(Relative atomic masses: Al = 27.0, Cl = 35.5, H =1) [rmc-29]

anonymous · Answer

$$89g AlCl_3 (\frac{1 mole AlCl_3}{223.5 g AlCl_3})(\frac{2 mole  Al}{1 mole AlCl_3})(\frac{27.0 g Al}{1 mole Al})=21.5 g Al$$

So you will need 21.5 grams of Al inorder to produce the 89g of AlCl3 that you needed.