PCl _ 5 ⇌ PCl _ 3 + Cl _ 2 K _ c = 1 . 844 3 . 0 moles of PCl _ 5...

${PCl}_{5 ( g )} ⇌ {PCl}_{3 ( g )} + {Cl}_{2 ( g )}{K}_{c} = 1 . 844t = 0300t = {t}_{eq}3 - xxx$

$⇒ \dfrac{\left[{PCl}_{3}\right]\left[{Cl}_{2}\right]}{\left[{PCl}_{5}\right]} = \dfrac{{x}^{2}}{3 - x} = 1 . 844$
$⇒ {x}^{2} + 1 . 844 - 5 . 532 = 0$
$⇒ x = \dfrac{- 1 . 844 + \sqrt{( 1 . 844{)}^{2} + 4 \times 5 . 532}}{2}$
$≡ 1 . 60$
$⇒$ Moles of ${PCl}_{5} = 3 - 1 . 60 = 1 . 40$