The reaction CO(g) + 2 H2(g)--> CH3OH(g) was studied using a 5.19 L flask and an initial condition of 26.9 g CO and 2.34 g H2. At equilibrium, the flask contains 8.34 g CH3OH. Calculate Kc for this reaction.

To calculate the equilibrium constant (Kc) for the given reaction, we need to know the concentrations of CO, H2, and CH3OH at equilibrium.

First, we need to convert the given masses (in grams) of CO, H2, and CH3OH to moles using their respective molar masses. The molar masses are as follows:
- CO: 28.01 g/mol
- H2: 2.02 g/mol
- CH3OH: 32.04 g/mol

Number of moles of CO:
moles of CO = mass of CO (g) / molar mass of CO (g/mol)
moles of CO = 26.9 g / 28.01 g/mol
moles of CO = 0.960 mol

Number of moles of H2:
moles of H2 = mass of H2 (g) / molar mass of H2 (g/mol)
moles of H2 = 2.34 g / 2.02 g/mol
moles of H2 = 1.16 mol

Number of moles of CH3OH:
moles of CH3OH = mass of CH3OH (g) / molar mass of CH3OH (g/mol)
moles of CH3OH = 8.34 g / 32.04 g/mol
moles of CH3OH = 0.260 mol

Now, we need to determine the concentrations (in moles per liter) of CO, H2, and CH3OH. The volume of the flask is given as 5.19 L.

Concentration of CO at equilibrium:
CO concentration = moles of CO / volume of flask
CO concentration = 0.960 mol / 5.19 L
CO concentration = 0.185 M

Concentration of H2 at equilibrium:
H2 concentration = moles of H2 / volume of flask
H2 concentration = 1.16 mol / 5.19 L
H2 concentration = 0.223 M

Concentration of CH3OH at equilibrium:
CH3OH concentration = moles of CH3OH / volume of flask
CH3OH concentration = 0.260 mol / 5.19 L
CH3OH concentration = 0.050 M

Now that we have the concentrations of CO, H2, and CH3OH at equilibrium, we can calculate Kc using the balanced chemical equation:

CO(g) + 2 H2(g) --> CH3OH(g)

Kc = [CH3OH]^1 / [CO]^1 * [H2]^2

Kc = (0.050) / (0.185) * (0.223)^2
Kc = 0.050 / (0.185 * 0.223^2)
Kc ≈ 0.502

Therefore, the value of Kc for this reaction is approximately 0.502.