1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    Yes.

    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

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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
    Thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Percent Yeild question.
  1. Theoretical Yeild (Replies: 2)

Loading...