Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:
2A1 + 3Cl2 + 2AlCl3
How many grams of aluminum chloride could be produced by
reacting 34.0 g of aluminum and 39.0 g of chlorine gas?

48.95g

Explanation:

no. of moles of Cl2 = 39/(2*35.5) = 0.55 mol

no. of moles of Al = 34/27 = 1.26 mol

hence, aluminium is in excess so we'll do calculation using no. of moles of Cl2 as it will be the only reactant to be used up completely. So,

no of moles of AlCl3 = 2/3 * (0.55) = 0.367 mol

hence amount of AlCl3 = 0.367 * (27+3*35.5) = 48.95g