Question

Find the % yield of cisplatin which is formed in the result of when 2 moles of ammonia react with potassium tetrachloro platinate

Determine the mass of nh3 that is needed to complete react with platinate and the amount of cisplatin that form?

An experiment begin with 20g of platinate

And 12.36 actual yield of cisplatin.

Answer 1

n(K₂PtCl₄) = 20 / 415 = 0.0482 mol

n(Pt(NH₃)₂Cl₂) = 12.36 / 300 = 0.0412 mol

$\alpha$ = 0.0412 / 0.0482 × 100% = 85.5%

m₂(NH₃) = 0.0482 × 17 = 0.82 g

m₂(Pt(NH₃)₂Cl₂) = 0.0482 × 300 = 14.46 g