Question

When 0.59 mol I2(g) and 0.59 mol H2(g) are placed into a 1.0-L container at a given temperature, 0.30 mol H2(g) is found to be present after the reaction below reaches equilibrium. Calculate Kc at the given temperature.

I2(g) + H2(g) 2HI(g)

Answer 1

You don't believe in arrows? How are we to know the reactants from the products?
........I2(g) + H2(g) ==> 2HI(g)
initial.0.59... 0.59......0
change...-x.....-x.........+2x
equil...........0.3

0.59-x = 0.3; therefore, x = 0.29 moles H2 used. That makes I2 = 0.59-0.3 = 0.29; HI must be 0 + 2x = 0.6 moles.
(HI) = moles/L
(H2) = moles/L
(I2) = moles/L
Substitute into Kc expression and solve for Kc.