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

## Aluminum burns bromine, producing aluminum bromide: $2 A l \left(s\right) + 3 B {r}_{2} \left(l\right) \to 2 A l B {r}_{3} \left(s\right)$ in a certain experiment, 6.0 g aluminum was reacted with an excess of bromine to yield 50.3g aluminum bromide.

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.

$\text{Moles of aluminum}$ $=$ $\frac{6.0 \cdot g}{26.98 \cdot g \cdot m o {l}^{-} 1} = 0.222 \cdot m o l$.

Had all the metal reacted, there would be $0.222 \cdot m o l \times 266.29 \cdot g \cdot m o {l}^{-} 1$ $=$ $59.12 \cdot g \text{ } A l B {r}_{3}$

Thus yield $=$ $\text{Recovered mass"/"Theoretical mass}$ $=$

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