Question

A piece of aluminum foil 1.00cm square and 0.590mm thick is allowed to react with bromine to form aluminum bromide. Part A How many moles of aluminum were used? (The density of aluminum is 2.699 g/cm^3 Part B How many grams of aluminum bromide form, assuming that the aluminum reacts completely?

Answer 1

Answer:

Part A: 5.899x10^-3 moles of Al

Part B: 1.573 g of AlBr3

Explanation:

Part A: We have to obtain the volume of the piece of aluminium; all sides of the square must be in cm. Then, use the density to obtain the mass.

$0.590 mm (\frac{1 cm}{10 mm}) = 0.059 cm$

$V= lxlxl= (1cm)(1cm)(0.059 cm)= 0.059 cm^3.$

0.059 is the volume of the Al udes for the reaction. Now, to oabtain the moles:

$0.059 cm^3 Al(\frac{2.699 g Al}{1 cm^3 Al})(\frac{1 mol Al}{27 g Al})= 5.899x10^-3 moles$

Part B: To obatin the mass of AlBr3, we need the balanced chemical equation:

                                               2Al + 3Br2 → 2AlBr3

We assume bromine (Br2) is in excess, therefore, we calculate the aluminum bromide formed from the Al:

$5.898 x 10^-3 moles Al (\frac{2 moles AlBr3}{2 moles Al}) (\frac{266.7 g AlBr3}{1 mol AlBr3}) = 1.573 g$ of Al