% yield of a reaction

If 54g of Aluminum is reacted with 480g of Bromine and 500g of Aluminum Bromide is recovered what is the % yield for the reaction?

2Al + 3Br2 = 2AlBr3

molecular mass of Al = 27 g/mole
molecular mass of Br2 = 160 g/mole
molecular mass of AlBr3 = 267 g/mole


no.of moles of Al = 54/27 = 2
no. of moles of Br2= 480/160 = 3

according to balanced reaction ...

when 2 moles of Al reacts with 3 moles of Br2 ....we get 2 moles of AlBr3

now,2 moles of AlBr3 weighs = 2 X 267 = 534 g

% yield = actual yield/theoretical yield X 100

where actual yield = 500 g
and theoretical yield = 534 g

so % yield = 500/534 X 100 = 0.93633 X 100 = 93.633%