Question

A piece of aluminum foil 1.00 cm square and 0.550 mm thick is dropped into bromine and reacts violently to form aluminum bromide solid. The density of aluminum is 2.699 g/cm³. A) Write a balanced equation for this reaction. B) If the reaction goes to completion, how many moles of aluminum reacted? C) How many grams of aluminum bromide formed, assuming the aluminum reacts completely?

Answer 1

Solution :-

A) Balanced reaction equation is as follows

2Al(s) + 3Br2(l) --------- > 2AlBr3 (s)

B) Now lets calculate the volume of the Aluminum foil (0.550 mm = 0.055 cm)

Volume = 1.00 cm2 * 0.055 cm = 0.055 cm3

now lets calculate the mass of the Al

mass = volume * density

         = 0.055 cm3 * 2.699 g per cm3

         = 0.1484 g Al

So moles of Al reacting = 0.1484 mol Al

C) Now lets calculate moles of the AlBr3 using the moles of the Al

mole ratio of the Al to AlBr3 is 2 :2

therefore moles of AlBr3 are same as moles of Al

so moles of AlBr3 = 0.1484 mol AlBr3

now lets calculate the mass of AlBr3

mass = moles* molar mass

mass of AlBr3 = 0.1484 mol * 266.694 g per mol

                      = 39.6 g AlBr3

Therefore mass of the AlBr3 that can be produced = 39.6 g AlBr3