Question

Cisplatin, an anticancer agent, is used for the treatment of solid tumours is prepared by the reaction of ammonia with potassium tetra chloroplatinate.

$K_{2}PtC{l}_4+2N{H}_3⟶\left[Pt\right(N{H}_3{)}_2]C{l}_2+2KCl$

Assume that $10.0gof{K}_{2}PtC{l}_4$ and $10.0gofN{H}_3$ are allowed to react. $(K=39,Pt=195,Cl=35.5)$

The number of moles of excess reactant which remains unreacted is :

• A. $0.024$
• B. $0.34$
• C. $0.54$
• D. $0.56$

Answer 1

The correct option is C $0.54$

We have

$K_{2}PtC{l}_4+2N{H}_3\to \left[Pt\right(N{H}_3{)}_2]C{l}_2+2KCl$

1 mole of $K_{2}PtC{l}_4$ reacts with 2 moles of $N{H}_3$

Given, mass of $K_{2}PtC{l}_4$ = 10g

molar mass of $K_{2}PtC{l}_4$ = 415

so, number of moles of $K_{2}PtC{l}_4$ = $\frac{10}{415}=0.024moles$

given mass of $N{H}_3$ = 10g

molar mass of $N{H}_3$ = 17

moles of $N{H}_3$ = $\frac{10}{17}=0.588moles$

so, 0.024 moles of $K_{2}PtC{l}_4$ reacts with 0.048 moles of $N{H}_3$

so, moles of $N{H}_3$ left unreacted = $0.588-0.048=0.54$