Question

Three moles of $P{Cl}_5$, three moles of $P{Cl}_3$ and two moles of ${Cl}_2$ are taken in a closed vessel. If at equilibrium the vessel has $1.5$ moles of $P{Cl}_5$, the number of moles of $P{Cl}_3$ present in it is:

• A. $5$
• B. $3$
• C. $6$
• D. $4.5$

Answer 1

The correct option is D $4.5$

	$PC{l}_5$	$PC{l}_3$	$C{l}_2$
Initial moles	$3$	$3$	$2$
Change in moles	$-x$	$x$	$x$
Equilibrium moles	$3-x=1.5$	$3+x$	$2+x$

Equilibrium moles of $PC{l}_5$ $=3-x=1.5$.

Hence, $x=1.5$

Equilibrium moles of $PC{l}_3=3+x=3+1.5=4.5$