When carbon monoxide reacts with hydrogen gas, methane and water vapor are formed according to...?

When carbon monoxide reacts with hydrogen gas, methane and water vapor are formed according to the following reaction
CO(g) + 3H2(g) = CH4(g) + H2O(g)
Analysis at equilibrium shows that Keq = 0.555, the [CO] = 0.921 M, the [H2] = 1.21 M, and the [CH4] = 0.0391 M. Calculate the equilibrium concentration for the water vapor.

Answer:
For a reaction aA + bB <====> cC + dD, the Keq is defined as
Keq = ([C]^c x [D]^d) / ([A]^a x [B]^b)

Use the formula for equilibrium

Keq = [CH4][H2O]/[H2]^2[CO]
So,
[H2O] = Keq * [H2] * [H2] * [CO] / [CH4]
[H2O] = 0.555*1.21*1.21*0.921/0.0391
[H2O] = 19.14 M

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