The equilibrium constant Kc equals 5.90 for the reaction

CH3COOH + C2H5OH <==> CH3COOC2H5 + H2O

Find the molar concentrations of acetic acid, ethanol, ethyl acetate, and water at equilibrium when .10 M acetic acid and .20 M ethanol are allowed to equilibrate.

I started like this:
I assumed 1 L of solution so I had
.10 mol CH3COOH
.20 mol C2H5OH

and in terms of moles at equilibrium...
CH3COOH= a - x
C2H5OH= b - x
CH3COOC2H5= c + x
H2O= d + x
where a, b, c, and d = moles initially, and x= moles lost

But I'm not really sure where to go from here. Any help would be much appreciated.

To find the molar concentrations of acetic acid, ethanol, ethyl acetate, and water at equilibrium, you will need to use the equilibrium constant expression and the given initial concentrations of acetic acid and ethanol.

First, let's assign variables to the initial concentrations and changes in concentrations as you have done:

a = initial moles of CH3COOH = 0.10 mol
b = initial moles of C2H5OH = 0.20 mol
c = initial moles of CH3COOC2H5 (ethyl acetate)
d = initial moles of H2O
x = moles lost (change in concentration at equilibrium)

Now, we need to express the equilibrium concentrations in terms of the initial concentrations and changes in concentrations:

CH3COOH = a - x
C2H5OH = b - x
CH3COOC2H5 = c + x
H2O = d + x

The next step is to write the equilibrium constant expression (Kc) using these equilibrium concentrations:

Kc = [CH3COOC2H5][H2O] / [CH3COOH][C2H5OH]

Since the equilibrium constant (Kc) is given as 5.90, we can substitute the equilibrium concentrations into this expression:

5.90 = ([c + x][d + x]) / ([a - x][b - x])

At this point, we have an equation with variables. To solve for x, we can rearrange the equation:

5.90 * ([a - x][b - x]) = ([c + x][d + x])

Expand the left side of the equation:

5.90 * (ab - ax - bx + x^2) = (cd + cx + dx + x^2)

Simplify:

5.90ab - 5.90ax - 5.90bx + 5.90x^2 = cdx + cx + dx + x^2

Now, move all the terms to one side of the equation:

5.90x^2 - (5.90a + 5.90b + c + d)x + (5.90ab - cd) = 0

This is a quadratic equation in terms of x. You can now solve for x using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 5.9, b = -(5.90a + 5.90b + c + d), and c = (5.90ab - cd). Plug in these values to solve for x.

Once you have solved for x, you can substitute the value back into the equilibrium concentration expressions to find the equilibrium concentrations of CH3COOH, C2H5OH, CH3COOC2H5, and H2O.