Question

Need help!! working on this for a while!!

Aluminum burns in liquid bromine according to the following equation: 2 Al(s) + 3 Br2(l) = 2 AlBr3(s) After reacting 20.0 g of aluminum, you isolate 87.4 g of aluminum bromide (AlBr3).

a). What was the theoretical yield for the reaction?

b). What was the percent yield for the reaction?

Answer 1

1) Balance the equation:

2 Al(s) + 3 Br2(l) = 2 AlBr3(s)

2) Find the limiting reagent : divide by Molecular wt. Al=26.98g/mol

20g/(26.98 g/mol) = .74 moles


3)Find the theoretical yield :multiply the amount of moles of the limiting reagent by the ratio of the limiting reagent and the synthesized product and by the molecular weight of the product.

2 moles Al -->2 moles Al Br3

.74 moles of Al (2molesAlBr3/2moles Al)(266.8g/mol AlBr3) = 197.34 grams

4) Actual yield : given was 87.4g

5)Percentage Yield = (Mass of Actual Yield / Mass of Theoretical Yield) ( x 100%)

87.4/197.34 =44.29%