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
anor277
Aug 15, 2016

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%

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