(27.0 g Al) / (26.98154 g Al/mol) = 1.00068 mol Al
(32.0 g Cl2) / (70.9064 g Cl2/mol) = 0.451299 mol Cl2
0.451299 mole of Cl2 would react completely with 0.451299 x (2/3) = 0.300866 mole of Al, but there is more Al present than that, so Al is in excess and Cl2 is the limiting reactant.
(0.451299 mol Cl2) x (2 mol AlCl3 / 3 mol Cl2) x (133.3405 g AlCl3/mol) = 40.1 g AlCl3
Answers & Comments
(27.0 g Al) / (26.98154 g Al/mol) = 1.00068 mol Al
(32.0 g Cl2) / (70.9064 g Cl2/mol) = 0.451299 mol Cl2
0.451299 mole of Cl2 would react completely with 0.451299 x (2/3) = 0.300866 mole of Al, but there is more Al present than that, so Al is in excess and Cl2 is the limiting reactant.
(0.451299 mol Cl2) x (2 mol AlCl3 / 3 mol Cl2) x (133.3405 g AlCl3/mol) = 40.1 g AlCl3