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$\pu{60g}$ of acetic acid and $\pu{46g}$ of ethyl alcohol are mixed with each other at a constant temperature and allowed to attain equilibrium. At equilibrium $\pu{58.2g}$ of ethylacetate and $\pu{12g}$ of water were present: find the concentration equilibrium constant ($K_\mathrm{c}$)

Here is my solution:

enter image description here

The correct answer is 4 but I'm getting 3.7 something. When I calculate x with the help of made of water it is slightly different than when I do it with the mass of ester. Shouldn't they both give the same answer? If not, which substance should I use?

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    $\begingroup$ In the given image, your $K_\mathrm{c}$ is indeed exactly equal to 4. (last step -> $K_\mathrm{c}=\frac{0.66^2}{0.33^2}=4$). Is there any other calculation page that you forgot to show us here where you're getting 3.7 as the answer? $\endgroup$ – Gaurang Tandon May 5 '18 at 1:21
  • $\begingroup$ No, the calculation is wrong. 1 - 0.66 = 0.34. Then Kc would be 3.7. The question is: can two products with the same molar ratio can have different concentrations at equilibrium? $\endgroup$ – Hammad Ahmed May 22 '18 at 4:13
  • $\begingroup$ 0.666 is actually the approximated value for 2/3. So, 1-0.66 is more accurately 1-2/3 which equals 1/3 which equals approximately 0.33 as expected. $\endgroup$ – Gaurang Tandon May 22 '18 at 4:21
  • $\begingroup$ The concentration of ethyl acetate at eq. is 0.661363636 and the concentration of water at eq. is 0.666666666. That's a slight difference but they give different values of Kc: 3.7 and 4 respectively. Is that an error in the given data that the two products with the same molar ratio have different concentrations at equilibrium or is there some rule by which I should prefer one product over the other for determining x which in turn helps calculates Kc? $\endgroup$ – Hammad Ahmed May 22 '18 at 4:29

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