Three moles of $\mathrm{PCl}_{5}$, three moles of $\mathrm{PCl}_{3}$ and two moles of $\mathrm{Cl}_{2}$ are taken in a closed vessel. If at equilibrium the vessel has $1.5$ moles of $\mathrm{PCl}_{5}$, the number of moles of $\mathrm{PCl}_{3}$ present in it is

• A

  5
• B

  3
• C

  6
• Correct answer
  D

  4.5

Solution

$\therefore$ At equilibrium $1.5$ moles of $\mathrm{PCl}$ is converted into $4.5$ moles of $\mathrm{PCl} 3$. $\therefore$ Total moles of $\mathrm{PCl}_{3}$ at equilibrium is $4.5$.

Hence, the correct answer is option: 'D'.