The reaction forms 98.76 g AlCl_3.
We have the masses of two reactants, so this is a limiting reactant problem.
We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.
Step 1. Gather all the information in one place with molar masses above the formulas and everything else below them.
M_r: ___26.98 _70.91 __133.34
________2Al + 3Cl_2 → 2AlCl_3
Mass/g: 20.00 _78.78
Step 2. Calculate the moles of each reactant
Moles of Al = 20.00 g Al × (1 mol Al /26.98 g Al) = 0.741 29 mol Al
Moles of Cl_2 = 78.78 g Cl_2 × (1 mol Cl_2 /70.91 g Cl_2) = 1.11 10 mol Cl_2
Step 3. Identify the limiting reactant
Calculate the moles of AlCl_3 we can obtain from each reactant.
From Al: Moles of AlCl_3 = 0.741 29 mol Al × (2 mol AlCl_3/2 mol Al) = 0.741 29 mol AlCl_3
From Cl_2: Moles of AlCl_3 = 1.11 10 mol Cl_2 × (2 mol AlCl_3/3 mol Cl_2) = 0.740 66 mol AlCl_3
Cl_2 is the limiting reactant because it gives the smaller amount of AlCl_3.
Step 4. Calculate the mass of AlCl_3.
Mass = 0.740 66 mol AlCl_3 × 133.34 g/1 mol AlCl_3 = 98.76 g AlCl_3
The reaction produces 98.76 g AlCl_3.