Question

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:

2Al(s)+3Cl2(g)?2AlCl3(s)

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

Answer 1

2 AICl3 (s) 2 Al (s) +3 Cl2 (g) No. of moles of Al = (27.0 g) / (27 g/mol) = 1 mol No. of moles of Chlorine = (32.0 g)/ (70.9 g/mol) = 0.45 mol No. of moles of Al needed to react with o.45 mol of Al (2 3) x 0.45-0.3 mol Here; Al present in excess So; the limiting reagent is C12 No. of mole sof AICI3 produced = (2 , 3) x 0.45 mol = 0.3 mol weight of AICI3 produced 0.3 mol) x (133.33 g/mol) = 40.1 grams