Percentage Yield

[Poojapriya]

Hi,
I need help with the following question:

15.3 grams of Lithium is dropped into a solution containing excess copper II phosphate. When the reaction is completed, 1.25 grams of copper is formed. What is the percent yield?

I worked the percent yield to be 1.8%, which I think isn't right. Can you please attach a picture of the steps required to get the percent yield? I've also attached the steps I have used below.

Thank you!

[sharlt]

Hi!

I'm not completely sure if this is right but this is what I got

2Li(s) + Cu3(PO4)2 -> 2LiPO4 (aq) + 3Cu(s)
N(Li) = 15.3/6.94 = 2.20 (2 moles) -> 1.10 (one mole)
So theoretically 3.30 moles (1.10 x 3) of Cu should form
N(Cu) = 1.25/63.55 = 0.02 moles
Percentage Yield = Actual yield/ Theoretical Yield = 0.02/3.30 x 100 = 0.6%