First, we determine the moles of aluminum and bromine present using their atomic masses and given masses:
moles = mass / atomic mass
Aluminum:
3 / 27 = 0.11 mole
Bromine:
6 / 80 = 0.075
The molar ratio of bromine to aluminum in the product is 3 : 1, so the moles of aluminum that will be needed are:
0.075 / 3 = 0.025
So, 0.025 moles of product will be formed as the reaction is limited due to bromine. This means the mass of product will be:
Mass = moles * Mr
Mass = 0.025 * 207
Mass = 5.18 grams of product
The remaining reagents will be:
Br - 0 grams as all will react
Al - (0.11 - 0.025) * 27 = 2.30 grams as it is in excess
If the reaction was to go to 72% yield, the mass formed would be 72% of the mass formed when the reaction completes. So:
Mass = 0.72 * 5.18
Mass = 3.73 grams
