Short Answer : Aluminum bromide can be prepared by the reaction of aluminum metal with bromine gas shown by the equation: 2 Al + 3 Br2 --> 2 AlBr3 Now suppose that 5.6 mol of aluminum reacts with 4.4 mol of bromine. 1. Calculate the mass of aluminum bromide that can be produced from 5.6 mol of Al. 2. Calculate the mass of aluminum bromide that can be produced from 4.4 mol of bromine. 3. Based off of your final answers for 1 and 2 above: The limiting reactant is _________________________. The excess reactant is __________________________. The actual amount of product produced is __________________________.