Question

You react 21.3 grams of dangerous aluminum metal with 66.5 grams of chlorine gas (Cl2) and get AlCl3 as the only product.

a) Which is the limiting reactant?

b) How much of the excess reactant will you have left over?

c) If your % yield is 75.4%, how much AlCl3 will you get?

Answer 1

2Al + 3Cl₂ ---------> 2AlCl₃

21.3 grams of aluminum = 21.3 / 26.98 = 0.78947 Mole

66.5 grams of chlorine = 66.5 / 35.45 = 1.8758 Mole

Cl2 is limitng reagent

1.8758 Mole of Cl2 will react with ( 1.8758 / 3) 0.62529 mole of aluminium .

excess of aluminum = 0.78947 - 0.62529 = 0.16417 Mole x 26.98 = 4.4294 gm Aluminum is unreacted

0.62529 mole of aluminium will produce the same mole AlCl3

0.62529 mole of AlCl3 = 0.62529 x 133.34 = 83.37 gm of AlCl3 for 100% yield

83.37 gm x 75.4 /100 = 62.86 gm of AlCl3 will get for 75.4 % yield.