The correct option is B 0.6
Here Kc=0.45.[Cl2]=0.15M
No. of moles of Cl2 at equrilibrium concentration is 0.15×3=0.45moles.
Let the no of moles of PCl5 added is 'x'
PCl5⇌PCl3+Cl2
Then,
(x−0.453) will be the equilibrium concentration of PCl5
∴Kc=[PCl3][Cl2][PCl5]=0.45
⇒0.15×0.15(x−0.453)=0.45
⇒x=0.6moles