how do you solve this??

Question

K_eq = 3.143x10^-5 for the reaction below:

NH2CH2COOH(aq) + C2H5OH(aq) <--> NH2CH2COOC2H5(aq) + H2O(l)

So if you were to combine 5.171g of NH2CH2COOH and 8.167g of C2H5OH in enough water to make 160.00mL of the solution, what are the equilibrium concentrations of all reactants and products?

Answer 1

Let's call the chemicals A, B, and C (the water is taken as irrelevant to the equilibrium)

K = [C]/([B][A])

You need to convert the grams of A and the grams of B by dividing by the appropriate molar masses and dividing both again by 0.160 liters in order to get [A]₀ and [B]₀ , the initial concentrations of the 2 reactants.

Letting x be the concentration of C produced in order to reach equilibrium, we obtain

k = 3.143 x 10^-5 = x/(([A]₀-x)([B]₀ -x) which can be solved using the quadratic formula

The final concentrations should be easy to calculate knowing x.

Please consider a tutor. Take care.