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Percent Yield

  1. Dec 28, 2010 #1
    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 %
     
  2. jcsd
  3. Dec 28, 2010 #2
    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.
     
  4. Dec 28, 2010 #3

    Borek

    User Avatar

    Staff: Mentor

    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

    conversion coefficient.

    --
    methods
     
  5. Dec 29, 2010 #4
    Okay, then it would be

    2.5 g Al x (1 mol Al / 27 g) x (2 mol Al2O3 / 4 mol Al)
     
  6. Dec 29, 2010 #5

    Borek

    User Avatar

    Staff: Mentor

    Much better now.
     
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