Question

Cisplatin [Pt(NH3)2Cl2], a compound used in cancer treatment, is prepared by reaction of ammonia with potassium tetrachloroplatinate:
K2PtCl4+2NH3→2KCl+Pt(NH3)2Cl2

How many grams of NH3 are needed to react with 60.1 g of K2PtCl4 ?

How many grams of cisplatin are formed from 60.1 g of K2PtCl4 if the percent yield for the reaction is 97 % ?

Answer 1

Answer:

K2PtCl4 + NH3 = 2KCl + Pt(NH3)2Cl2

n(K2PtCl4) = m/M

n(K2PtCl4) = 60.1/415 = 0.145 mol

n (NH3) = 0.290 mol

m (NH3) = 0.290*17 = 4.92 g

n (Pt(NH3)2Cl2) = 0.145 mol

m (Pt(NH3)2Cl2) = 0.145*300 = 43.5 g

97% = x/43.5

x = 42.195 g