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Percent Yeild question.

  1. Jul 7, 2009 #1
    1. The problem statement, all variables and given/known data

    Given the following equation:

    K2PtCl4 + 2 NH3 -----> Pt(NH3)2Cl2 + 2 KCl

    c) Starting with 34.5 g of NH3, and you isolate 76.4 g of Pt(NH3)2Cl2, what is the percent yield?

    3. The attempt at a solution

    I had to balance the eqn, which was pretty easy. Then I determined what the limiting reagent is, given 34.5g NH3 (which was asked) and I deduced that NH3 was the LR and calculated the theoretical yeild of KCl to be 151g. Part C of the question is what I am having some difficulty with. Percentage yeild.

    I'm thinking that the plan for this problem is to calculate the moles of each and then just turn those numbers into percentages. But I don't think that's right. I kind of don't really know where to start otherwise. I could calculate the moles of each and then take a percentage by dividing moles of NH3 by the moles of Pt(NH3)2Cl2 and multiplying that by 100 to make a percentage figure. Need help on this one guys.
  2. jcsd
  3. Jul 7, 2009 #2
    If you know the LR then you can figure out the theoretical yield of both products, correct?

    Percent yield is just (actual yield / theoretical yield)
  4. Jul 7, 2009 #3

    So lets say I have the theoretical yeild of KCl and Pt(NH3)2Cl2, which is easy to calculate. What is the "actual yield"? When I do the calculation of (actual yield / theoretical yield) what figures am I putting in?
  5. Jul 7, 2009 #4


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    Wasn't the actual Pt(NH3)2Cl2 yield given as 76.4 g? So you just need to calculate the theoretical yield of Pt(NH3)2Cl2, given 34.5 g of NH3.
  6. Jul 7, 2009 #5
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