In an experiment, 8.0 moles of H2(g) is heated with 5.0 moles of I2(g) in a closed vessel at 298 K. At equilibrium, 9.06 mol of hydrogen iodide are formed. Calculate the equilibrium constants Kc and Kp at that temperature (R = 8.314 J mol-1 or 0.082 atm dm3 K-1 mol-1).

The reaction is between hydrogen gas and iodine gas to form hydrogen iodide gas.

Initial moles: $\text{H}_2 = 8.0 \text{ mol}$, $\text{I}_2 = 5.0 \text{ mol}$, $\text{HI} = 0 \text{ mol}$.
Let $x$ be the moles of $\text{H}_2$ and $\text{I}_2$ that react. According to the stoichiometry, $2x$ moles of $\text{HI}$ will be formed.

At equilibrium, $9.06 \text{ mol}$ of hydrogen iodide ($\text{HI}$) are formed.
So, $2x = 9.06 \text{ mol}$.

Equilibrium moles of $\text{H}_2 = 8.0 - x = 8.0 - 4.53 = 3.47 \text{ mol}$
Equilibrium moles of $\text{I}_2 = 5.0 - x = 5.0 - 4.53 = 0.47 \text{ mol}$
Equilibrium moles of $\text{HI} = 9.06 \text{ mol}$

The expression for $K_c$ is:

Let the volume of the closed vessel be $V \text{ dm}^3$. The concentrations are moles divided by volume.

Substitute these into the $K_c$ expression:

The volume $V^2$ terms cancel out:

Rounding to three significant figures, $K_c = \boxed{\text{50.3}}$.

The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n}$.
$\Delta n$ is the change in the number of moles of gaseous products minus the number of moles of gaseous reactants.
For the reaction $\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2\text{HI(g)}$:
Moles of gaseous products = 2 (from 2HI)
Moles of gaseous reactants = 1 (from $\text{H}_2$) + 1 (from $\text{I}_2$) = 2
So, $\Delta n = 2 - 2 = 0$.
Since $\Delta n = 0$, $(RT)^{\Delta n} = (RT)^0 = 1$.
Therefore, $K_p = K_c \times 1 = K_c$.

Rounding to three significant figures, $K_p = \boxed{\text{50.3}}$.