Equilibrium constant calculation from concentrations
Well, I haven't done that in a while, but we seem to agree that the equilibrium constant is expressed as

$$ K_c = \frac{[AB]^2}{[A_2][B]^2}$$

What else do we know?


Initially, $[A_2]$ = 8.0M, $[B]$ = 10.0M
In equilibrium, $[AB]$ = 8.0M
For every mole of $B$, one mole of $AB$ is produced.
For every mole of $A_2$, two moles of $AB$ are produced.


With just one cup of coffee so far, I get

$$ K_c = \frac{8^2}{(8 - \frac{8}{2})(10-8)^2} = 4$$