The equilibrium constant Kc for the equation2H2 (g) + CO (g) ⇌ CH3OH (g) is 35 at a certain temperature. If there are 3.21 * 10-2 moles of H2 and 4.87 * 10-3 moles of CH3OH at equilibrium in a 3.63-L flask, what is the concentration of CO?

Answer: the concentration of [CO] = 0.0532M

Explanation:

From The equation of reaction

2H2 (g) + CO (g) ⇌ CH3OH (g)

Applying Kc = [CH3OH]/[H2]^2[[CO]

[CH3OH] = 0.00487

[CO] = x-0.00487

[H2] = (0.032-0.00487) ^2=0.0271

Substitute into formula

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

35 = 0.00487 / (0.0271) ^2 (x-0.00487)

Simplify

x-0.00487=0.189

x = 0.00487+0.189=0.193moles

[CO] = n/C = 0.193/3.63 = 0.0532M