Calculating Percent Yield: K2PtCl4 + 2 NH3 -> Pt(NH3)2Cl2 + 2 KCl

[Attraction]

Homework Statement

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?

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.

 

[Feldoh]

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)

 

[Attraction]

If you know the LR then you can figure out the theoretical yield of both products, correct?

Yes.

Percent yield is just (actual yield / theoretical yield)

So let's 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?

 

[Redbelly98]

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.

 

[Attraction]

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.

Thanks.