What Is the Percent Yield of Al2O3 in This Reaction?

[priscilla89]

Homework Statement

If the reaction of 2.5 g of Al with 2.5 g of O2 produced 3.5 g of Al2O3, what is the percent yield of Al2)3? Be sure to write and balance equation.

Homework Equations

Percent Yield = (actual yield / theoretical yield ) x 100

The Attempt at a Solution

4Al + 302 ---> 2Al2O3

2.5 g Al x (1 mol Al / 108 g) x (2 mol Al2O3 / 4 mol Al) = .0115 mol Al2O3

.0115 mol Al2O3 (204 g Al2O3 / 1 mol Al2O3) = 2.36 Al2O3

Percent Yield = (2.36 / 3.50) X 100 = 67 %

 

[p21bass]

Two things:

1: Don't forget to verify which reagent (Al or O₂) is the limiting reagent.
2: Your conversion factors are wrong. For instance - there aren't 108g of Al in 1 mol of Al.

 

[Borek]

108g is a mass of 4 moles of Al, not of 1 mole. 4 is a stoichiometric coefficient, and it is already - correctly - present in your

(2 mol Al2O3 / 4 mol Al)

conversion coefficient.

--
methods

 

[priscilla89]

Two things:

1: Don't forget to verify which reagent (Al or O₂) is the limiting reagent.
2: Your conversion factors are wrong. For instance - there aren't 108g of Al in 1 mol of Al.

Okay, then it would be

2.5 g Al x (1 mol Al / 27 g) x (2 mol Al2O3 / 4 mol Al)

 

[Borek]

Much better now.