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priscilla89 Dec28-10 11:15 AM

Percent Yield
 
1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. 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 Dec28-10 03:01 PM

Re: Percent Yield
 
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 Dec28-10 03:30 PM

Re: Percent Yield
 
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

Quote:

Quote by priscilla89 (Post 3056337)
(2 mol Al2O3 / 4 mol Al)

conversion coefficient.

--
methods

priscilla89 Dec29-10 06:54 AM

Re: Percent Yield
 
Quote:

Quote by p21bass (Post 3056619)
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 Dec29-10 08:08 AM

Re: Percent Yield
 
Much better now.


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