Question

How many grams of Bromine are required to react completely with 11.95 grams aluminum chloride? AlCl3+Br2-->AlBr3+Cl2

Answer 1

First have to balance the equation of AlCl3 + Br2 ---> Br3 + Cl2:

2AlCl3 + 3Br3----> 2AlBr3 + 2Cl2

1 mol AlCl3 3mol Br2 159.8 g Br2 11.95g AlCl3 x --------------- x ------------- x --------------- = 21.48g Br2
133.33 g AlCl3 2 mol AlCl3 1 mol Br2


(133.33 g = molar mass of AlCl3 3:2 ratio
159.8 g = molar mass of Br2)

Answer: 21.48 g Br2 required

Answer 2

Molar mass:

( Br₂ ) = 159.80 g/mol

AlCl₃ = 133.34 g/mol

2 AlCl₃ + 3 Br₂ = 2 AlBr₃ + 3 Cl₂

2 x ( 133.34) g ------------ 3 x ( 159.80) g
11.95 ------------------------ ( mass Bromine )

mass Bromine = 11.95 x 3 x 159.80 / 2 x 133.34

mass Bromine = 5728.83 / 266.68

mass Bromine = 21.48 g

hope this helps!