Answer
See the answer below.
Work Step by Step
$n_{COBr_2,Reacted}=x, [COBr_2]_{equil}=(0.0500-x)/2.00$
$[CO]_{Equil}=x/2.00, [Br_2]_{Equil}=x/2.00$
Equilibrium constant:
$K=[CO][Br_2]/[COBr_2]$
$0.190=x/2\times x/2\div (0.0500-x)/2$
$0.190\times (0.0500-x)/2\times4=x^2$
$0.019-0.38x=x^2$
$x^2+0.38x-0.019=0$
$x=0.0447\ mol$
At equilibrium:
$[CO]=0.0224\ M=[Br_2], [COBr_2]=2.63\times10^{-3}\ M$
Fraction decomposed of COBr2
$f=x/0.05=0.954=95.4\%$
