Answer
See the answer below.
Work Step by Step
Initial concentration:
$[PCl_5]=3.120\ g\div 208.24\ g/mol\div 10.0\ L=0.0015\ M$
$[PCl_3]=3.845\ g\div 137.33\ g/mol\div 10.0\ L=0.0028\ M$
$[Cl_2]=1.787\ g\div 70.91\ g/mol\div 10.0\ L=0.0025\ M$
Added:
$[Cl_2]=1.418 g\div 70.91\ g/mol\div 10.0\ L=0.0020\ M$
$K=[PCl_3][Cl_2]/[PCl_5]$
$K=0.0025\times0.0028/0.0015=0.0047$
The equilibrium will shift to the left, so concentration of reacted: x
$0.0047=(0.0028-x)(0.0045-x)/(0.0015+x)$
$x^2-0.012x+5.55\times10^{-6}=0$
$x=0.00048$
$[PCl_5]=0.0020\ M$
$[PCl_5]=0.0023\ M$
$[Cl_2]=0.0040\ M$
