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?

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.

To find the number of moles of HI(g) present when the system reaches equilibrium, we need to apply the concept of the equilibrium constant and the initial conditions.

First, let's define the equilibrium constant expression for the given reaction:

Kc = [I2(g)][H2(g)] / [HI(g)]^2

Given that the equilibrium constant (Kc) is 0.020, we can set up the following equation:

0.020 = [I2(g)][H2(g)] / [HI(g)]^2

Now, let's consider the initial conditions. Initially, we have 1.00 mol of HI(g) in a 0.750-L container. Therefore, the initial concentration of HI is:

[HI]initial = moles of HI / volume of container
[HI]initial = 1.00 mol / 0.750 L
[HI]initial = 1.33 M (Molar concentration)

Now, let's assume that "x" moles of HI react to produce I2(g) and H2(g), according to the balanced equation. Therefore, the change in concentration of HI will be -x, while the changes in concentrations of I2 and H2 will be +x each.

At equilibrium, the equilibrium concentrations will be given by:

[HI]equilibrium = [HI]initial - x
[I2]equilibrium = x
[H2]equilibrium = x

Now, substituting the equilibrium concentrations into the equilibrium constant expression, we have:

0.020 = [I2(g)][H2(g)] / [HI(g)]^2
0.020 = (x)(x) / ([HI]initial - x)^2

Simplifying the equation, we get:

0.020 = x^2 / (1.33 - x)^2

Now, we can solve this equation to find the value of "x", which represents the number of moles of HI(g) that will be present when the system reaches equilibrium.

However, since this is a quadratic equation, it might require solving using numerical methods such as trial and error, or by using a graphing calculator or software. Therefore, the exact value of "x" cannot be determined without further calculation.

Note: In this case, the given equilibrium constant and the initial conditions enable us to set up the equation for solving the equilibrium concentration of HI(g). However, the actual value of "x" would require additional calculations based on the specific conditions.