Answer
See the answer below.
Work Step by Step
$n_{I_2,Reacted}=x, [I_2]_{Equil}=(0.105-x)/12.3$
$n_{I,Formed}=2x, [I]_{Equil}=2x/12.3$
Equilibrium constant:
$K=[I]^2/[I_2]$
$3.76\times10^{-3}=(2x/12.3)^2/[(0.105-x)/12.3]$
$3.76\times10^{-3}\times(0.105-x)/12.3\times(12.3/2)^2=x^2$
$x^2=0.00121-0.0116x$
$x^2+0.0116x-0.00121=0$
$x=0.0295$
At equilibrium:
$[I_2]=0.0061\ M, [I]=0.0048\ M$
