Question

The following reaction has an equilibrium constant of 0.020 at a given temperature.

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

If you have 1.00 mol HI(g) in a 0.750-L container initially, how many moles of HI(g) will be present when the system reaches equilibrium?

Answer 1

Technically you don't have an equation because you omitted the arrow. You MUST include an arrow for us to know where the reactants stop and the products start.
2HI ==> H2 + I2

K = 0.020 = (H2)(I2)/(HI)^2
Set up an ICE CHART and solve.
initial:
H2 = O
I2 = 0
HI = 1.00 mole/0.750L = 1.33

change:
H2 = +x
I2 = +x
HI = -2x

equilibrium:
H2 = +x
I2 = +x
HI = 1.33-2x

Substitute into the K expression and solve for x.