In an experiment, 6.67 g of AlCl3 was produced and 0.54g A1 remained unreacted. How many g atoms of Al and Cl2 were taken originally (Al =27, Cl = 35.5)?

Moles of AlCl3 produced = 6.67133.5=0.05molExcess of Al = 0.5427=0.02moleg atom or moles of Al taken = 0.05+0.02 = 0.07g atom or moles of Cl2 taken = 3 x 0.05 = 0.15