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Homework Help: Percent yield crazy--help with calculations?

  1. Oct 21, 2017 #1
    1. The problem statement, all variables and given/known data

    Consider:

    2 mol Reactant A (aq) + 3 mol Reactant B (aq) = 6 mol Product A (aq) + 1 mol Product B (s)

    You have 0.00075 L of a 0.0514M solution of Reactant A. In the lab, you combine it with 0.042 L of 0.0753M of Reactant B. You get a solid product and an aqueous product (a precipitation reaction!). Please, find the limiting reagent. Then consider this: if you recover 0.0569 g of Product B, then what is your percent yield? Pretend Product B's weight is 310.18g/mol.

    2. Relevant equations
    N/a

    3. The attempt at a solution

    Finding the Limiting Reagent:

    Considering Reactant A
    [(7.5*10^-4 L Reactant A/x moles Reactant A] = [(1 L Reactant A)/0.054 moles Reactant A]= 3.855*10^-5 moles reactant A used

    [3.855*10^-5 moles reactant A/x moles Product B]=[2 moles Reactant A/1 mole Product B]= 1.9275*10^-5 moles product B made

    Considering Reactant B
    [0.042L Reactant B/x moles Reactant B]=[1 L Reactant B/0.0753 moles Reactant B]= 0.0032 moles Reactant B used

    [0.0032 moles Reactant B/ x moles Product B]=[3 moles Reactant B/1 mole Product B]= 0.0011 moles Product B made

    :.Limiting Reagent is Reactant A.

    Yields:
    Theoretical yield:
    1.9275*10^-5 moles product B *(310.18g/mol)= 0.006 g Product B

    Percent Yield: (0.0569g/0.006g)*100 = 948.3%

    Well. That doesn't seem right, but I can't find where I went wrong.
     
  2. jcsd
  3. Oct 21, 2017 #2

    mfb

    User Avatar
    2017 Award

    Staff: Mentor

    That is a very strange notation.

    I get the same result as you, I guess the problem statement has a factor 10 wrong somewhere.

    You shouldn't round the theoretical yield to one digit, by the way, it introduces a large rounding error.
     
  4. Oct 22, 2017 #3

    Borek

    User Avatar

    Staff: Mentor

    Yep, must be an error in the problem statement.
     
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