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]`
The number of mol of `K_(2) [PtCl_(4)]` and `NH_(3)` used, respectively, are
A. 0.1,0.2
B. 0.2,0.04
C. 0.3,0.6
D. 0.03,0.06

Answer 1

Correct Answer - B
Mol of `K_(2) [PtCl_(4)]` consumed `= 0.2 "mol"`
= mol of cisplatin
`NH_(3)` consumed `= 2 xx 0.2 = 0.4 "mol"`.