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Basic Stoichiometry Question

  1. Dec 10, 2009 #1
    1. The problem statement, all variables and given/known data

    4Al + 3O2 --->2Al2O3

    Given 20g of Al and 15g Oxygen which is limiting factor?

    2. Relevant equations

    My question is the difference between O2 and Oxygen. Do I start with a mole of O or O2 in my calcs? Since the question is framed "15g of oxygen" (verbatim from a quiz) should it be inferred it's O2 in which case it's 32g/mol as opposed to 16g/mol, either .47 moles O2 or .94 moles O....?

    3. The attempt at a solution
    I used the O 16g/mol calc.
    15g O2 x 1(mole O)/16 (mm O) x 2(mol Al2O3)/3(mol O) = 0.625 moles Al2O3 produced
    20g Al x 1(mole Al)/27 (mm Al) x 2 (mol Al2O3)/4 (mol Al) = 0.37 moles AlsO3 produced

    With this approach Al is limiting factor. Professor briefly touched upon this problem and I had not chance to question. he was clear that O was limiting factor.

    is it my O vs. O2 calculations?

    Should my O calc be:

    15g O2 x 1(mol O)/32 (mm O2) x 2(mol Al2O3)/3(mol O) = 0.312 moles Al2O3 produced.


  2. jcsd
  3. Dec 11, 2009 #2


    User Avatar

    Staff: Mentor

    While 15 g of oxygen is 0.47 moles of O2 and 0.94 moles of O, it is still the same number of atoms (that is, amount of oxygen has not changed). However, your reaction is balanced with O2 - so you should use moles of O2 for calculation. If you use reaction balanced with O (that is, 2Al + 3O -> Al2O3) you use moles of O. That's not reasonable approach as oxygen is diatomic, but it will still give correct result.

    And yes, Al is a limiting reagent.

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