Question

Starting with 0.280mol SbCl3 and 0.160mol Cl2, how many moles of SbCl5, SbCl3, and Cl2 are present when equilibrium is established at 248?C in a 2.50?L flask? SbCl5(g)?SbCl3(g)+Cl2(g) Kc=0.025 at 248?C

I have the answers but can someone please show the work.

Answers:

SbCl5=.11587 mol
SbCl3=.16413 mol
Cl2= .04413 mol

Answer 1

concentration = moles /volume

[ SbCl3]= 0.28/2.5= 0.112M, [Cl2]=0.16/2.5=0.064

SbCl5(g)< ---------------------> SbCl3(g)+Cl2(g)

       0                                       0.112       0.064            at initial

     x                                        0.112 -x      0.064 -x      at equilibrium

Kc= [ SbCl3][Cl2]/[SbCl5]

0.025 = [0.064 -x][ 0.112 -x]/x

if we simplify this we can get x = 0.046

[SbCl5]= x= 0.046M

[SbCl5] in moles= Molarity x volume= 0.046x2.5=0.115 mole

[SbCl3]= 0.112 -x= 0.112-0.046=0.066M

[SbCl3] in moles= 0.066x2.5=0.165mole

[Cl2] = 0.064 -x= 0.064-0.046= 0.018M

[Cl2] in moles = 0.018x2.5=0.045mole