(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

2Al + 3Cl₂ → 2AlCl₃

2Al = 2•27 = 54

3Cl₂ = 6•35.5 = 213

2AlCl₃ = 2•27 + 6•35.5 = 267

54 g of Al + 213 g of Cl₂ → 267 g of AlCl₃

reacting 27.0 g of aluminum with 32.0 g of chlorine?

ratio Al to Cl₂ is is 54:213 so the 32 g is the determining factor

ratio Cl₂ to to AlCl₃ is 213:267 so 32 g of Cl₂ produces

32(267/213) = 40.1 g of AlCl₃

Aluminum reacts with chlorine....

Limiting reactant problem. Use each reactant to compute the amount of product. The lesser amount is the theoretical yield and tells the limiting reactant.

2Al(s) + 3Cl2(g) --> 2AlCl3(s)

27.0g .... 32.0g ...........?g

27.0g Al x (1 mol Al / 27.0g Al) x (1 mol AlCl3 / 1 mol Al) x (133.35g AlCl3 / 1 mol AlCl3) = 133.35g AlCl3

32.0g Cl2 x (1 mol Cl2 / 70.90g Cl2) x (2 mol AlCl3 / 3 mol Cl2) x (133.35g AlCl3 / 1 mol AlCl3) = 40.12g AlCl3

40.1g AlCl3 is produced and Cl2 is the limiting reactant.