The reaction forms 98.76 g AlCl_3.
We have the masses of two reactants, so this is a <em>limiting reactant problem</em>.
We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.
<em>Step 1. Gather all the information</em> 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
<em>Step 2</em>. Calculate the <em>moles of each reactant</em>
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 <em>limiting reactant</em>
Calculate the moles of AlCl_3 we can obtain from each reactant.
<em>From Al</em>: Moles of AlCl_3 = 0.741 29 mol Al × (2 mol AlCl_3/2 mol Al) = 0.741 29 mol AlCl_3
<em>From Cl_2</em>: Moles of AlCl_3 = 1.11 10 mol Cl_2 × (2 mol AlCl_3/3 mol Cl_2) = 0.740 66 mol AlCl_3
<em>Cl_2 is the limiting reactant</em> because it gives the smaller amount of AlCl_3.
<em>Step 4</em>. Calculate the <em>mass of AlCl_3</em>.
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.
