2HI(g)--H2(g) + I2(g) k eq= 8.0

Question

2.0 mol of HI are placed in a 4.0L container, and the system is allowed to reach equilibrium. Calculate the equilibrium concentration of all three gases
Initial
2HI=0.5 mol/L
H2= 0
I2=0
change
2HI= -2x
H2= x
I2=x
equilibrium
2HI= 0.5-2x
H2= x
I2=X
k eq= product/reactants
8.0= (x)(x)/ 0.5-2x
2.8=x/0.5-2x
x=0.21
so 0.21 mol/L of H2 and I2
HI= 0.42 0.5-2(0.21) = 0.08mol/L