Answer
See the answer below.
Work Step by Step
$n_{N_2O_4,Reacted}=x, [N_2O_4]_{Equil}=(15.6\ g/92.01\ g/mol-x)/5.000\ L$
$n_{NO_2,Formed}=2x, [NO_2]_{Equil}=2x/5.000$
Equilibrium constant:
$K=[NO_2]^2/[N_2O_4]$
$170=(2x/5.000)^2/[(0.170-x)/5.000]$
$170\times(0.170-x)/5\times(5/2)^2=x^2$
$x^2=36.13-212.5x$
$x^2+212.5x-36.13=0$
$x=0.1694\ mol$
At equilibrium:
$n_{NO_2}=0.339\ mol$
$f=(x)\div(15.6/92.01)=0.9992=99.92\%$
