Question

A piece of aluminum foil 0.536 mm thick and 1.21 cm square is allowed to react with bromine to form aluminum bromide.

(a) How many moles of aluminum were used? (The density of aluminum is 2.699 g/cm3.)
(b) How many grams of aluminum bromide form, assuming that the aluminum reacts completely?

Answer 1

Al + Br --> AlBr

a) volume of aluminium = 0.536*10^-2 cm * 1.21 cm^2 = 0.64856 cm^3

density of aluminium = 2.699 g/cm3

so mass = density*volume = 2.699*0.64856 = 1.75 gm

b) 1 mole of Al produces 1 mole of AlBr.

1 mole of AlBr has mass = 106.88 gm

Answer 2

0.536 mm = 0.0536 cm.

So the volume of Al is:

0.0536 * 1.21 * 1.21 = 0.0785 cm^3.

Using the density of Al, the mass is then:

2.699 g/cm^3 * 0.0785 cm^3 = 0.2118 g Al.

The atomic mass of Al is 26.98 g/mol, so this is equivalent to:

0.2118/26.98 = 0.00785 moles Al.

Now, 1 mole of Al can produce 1 mole of AlBr3. The MW of AlBr3 is 266.7 g/mol, so this amount of aluminum makes:

0.00785 * 266.7 = 2.09 g AlBr3.