how much aluminum chloride will be produced?

Question

Aluminum reacts with magnesium chloride to form aluminum chloride and magnesium metal. If 45.56 grams of aluminum react with 86.9 g of magnesium chloride, how much aluminum chloride will be produced?

Answer 1

2Al(s) + 3MgCl2(aq) ==> 2AlCl3(aq) + 3Mg(s) <-- Balanced equation

Determine which reactant is limiting:

moles Al present = 45.56 g Al x 1 mol Al/26.98 g = 1.688 moles Al

moles MgCl2 present = 86.9 g MgCl2 x 1 mol MgCl2/95.21 g = 0.909 moles MgCl2

Limiting reactant is MgCl2 based on mole ratio of 2 Al : 3 MgCl2 and 1.688 Al : 0.909 MgCl2

 

Now you can determine moles of product (AlCl3) will be produced:

0.909 moles MgCl2 x 3 moles Mg/3 moles MgCl2 = 0.909 moles AlCl3 produced  <-- Answer in moles

0.909 moles AlCl3 x 133.3 g/mole = 121 g AlCl3 produced  <-- Answer in grams