The reaction of bromine with aluminum chloride will bring about the formation of aluminum bromide and chlorine through a single-replacement reaction. This can be expressed in a balanced chemical reaction,
3Br₂ + 2AlCl₃ --> 2AlBr₃ + 3Cl₂
This means that 3 moles of Br₂ which is equal to 159.1 g is required to react with 2 moles of AlCl₃ or equal to 266.68 g. Using this and the given above,
mass of bromine = (37.4 grams AlCl3)(159.1 g of Bromine/266.68 g of AlCl3) Mass of bromine = 22.31 grams of Br₂
