Question

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.

Answer 1

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%`