Equilbrium constant question IB

answer is D

Is this missing any information? I thought you needed concentration to work out Kc?

Unless its just:
=[0.80]2 / 0.1*0.1

well yes it could be solved this way but how did you conclude 0.1?
is it by 0.80-2x
x= 0.40
i was confused if i should do that

Yeah what i thought is because its an equilibrium mixture where 0.80 mol of Ibr has formed, that means only 0.4 mol of each reactant is used.

so at equilibrium in the question we have 0.1 mol of each reactant remaining with 0.8 mol of the mixture - so I used those values for the equation.