Question

Cisplation is used an anticancer agent for the treatment of solid tumors, and its prepared as follows:
`underset({:("Potassium tetra"),("Chloro platinate"):})(K_(2)[PtCl_(4)]) + 2NH_(3) rarr underset("Ciplatin")([Pt(NH_(3))_(2) Cl_(2)]) + 2KCl`
Given `83.0 g` of `K_(2) [PtCl_(4)]` is used with `83.0 g` of `NH_(3)`.
Atomic weights: `K = 39 , Pt = 415, Cl = 35.5 N = 14]`
Which reactant is the limiting reagent and which is in excess?
A. `K_(2)[PtCl_(4)]` `NH_(3)`
B. `NH_(3)` `K_(2) [PtCl_(4)]`
C. None None
D. Both Both

Answer 1

Correct Answer - A
`Mw "of" K_(2)[PtCl_(4)] = 2 xx 39 + 195 + 4 xx 35.5 = 415 g`,
`Mw "of" NH_(3) = 17 g`
Mol of `K_(2) [PtCl_(4)] = (83.0)/(415) = 0.2` mol (limiting reagent)
Mol of `NH_(3) = (83)/(17) = 4.88 "mol"` (excess)