What is the maximum mass of aluminium chloride that can be formed when reacting 27.0 of aluminum with 32.0 of chlorine?

The answer is 40.1 g AlCl3, assuming that the questioner meant 27.0 g and 32.0 g (the units weren't provided in the question). It was calculated as follows: 27.0 g of Al is 27.0 g/27.0 g/mol = 1.00 mol of Al.

32.0 g of Cl is 32.0 g/35.45 g/mol = 0.903 mol of Cl.

Aluminium chloride has the molecular formula of AlCl3, therefore it takes three moles of chlorine to form one mole of aluminum chloride, or in this case, it would take 3.00 mol of Cl to react with 1.00 mol of Al. It is clear that the amount of AlCl3 that can be formed is limited by the moles of Cl available, specifically the maximum number of moles of AlCl3 that can be formed is 0.903 mol/3 = 0.301 mol.

0.301 mol AlCl3 is equivalent to: 0.301 mol AlCl3 x 133.35 g AlCl3/mol = 40.1 g AlCl3.