How do you calculate the theoretical and percent yields for this experiment?

Aluminum burns bromine, producing aluminum bromide:

#2Al (s) + 3Br_2 (l) -> 2AlBr_3 (s)#

in a certain experiment, 6.0 g aluminum was reacted with an excess of bromine to yield 50.3g aluminum bromide.

1 Answer
Aug 15, 2016

Answer:

Aluminum reduces bromine to give the salt in approx. 85% yield.

Explanation:

From your stoichiometrically balanced equation you note the 1:1 equivalence between moles of metal, and moles of salt: one mole of aluminum yields one mole of alumium tribromide given 100% yield.

#"Moles of aluminum"# #=# #(6.0*g)/(26.98*g*mol^-1)=0.222*mol#.

Had all the metal reacted, there would be #0.222*molxx266.29*g*mol^-1# #=# #59.12*g" "AlBr_3#

Thus yield #=# #"Recovered mass"/"Theoretical mass"# #=#

#(50.3*g)/(59.12*g*mol^-1)xx100%=85%#