Question

In the Reaction 2Al(s)+3Cl2(g)-> 2AlCl3(s)

a. Which reactant is the limiting reactant if 2.70mg Al and 4.05mg Cl2 react?

b. How many milligrams of AlCl3 can be produced from the conditions in part a?

Answer 1

Molecular masses:

26,98 Al 70,91 Cl2 133,34 AlCl3

a)

2.70mg Al equal to 0.1 mmol. It requires 3/2*0.1 = 0.15 mmol Cl2 or 0.15*70.91=10.64 mg. 4.05 mg only is given, so Cl2 is limiting!

b)

4.05mg/70.91mg/mmol = 0.057 mmol

it produces 0.057*2/3 = 0.038 mmol AlCl3

or m = 0.038 mmol * 133.34 = 5.08 mg