C₂H₅OH + CH₃COOH ---------------> C₂H₅CH₃COO + H₂O

INITIAL MOLES:                                 1              1                                      0              1

EQUILIBRIUM MOLES :           1-(54.3*1)/100    1-(54.3*1)/100        0+(54.3*1)/100    1+(54.3*1)/100

* SIGN DENOTES MULTIPLICATION

SINCE THEY ARE TELLING THAT ONLY 54.3% OF ACID IS CONVERTED THIS MEANS THAT 54.3% OF INITIAL MOLES IS CONVERTED INTO ESTER MEANS  (54.3*1)/100 MOLES IS REMOVED . AND THEN ACCORDING TO STICHIOMETERY SINCE ALL THE COEEFICIENTS ARE 1 THEREFORE THE NO. OF MOLES OF ACID WHICH IS CONVERTED INTO ESTER THAN MANY MOLES WILL ALSO BE REMOVED FROM  C₂H₅OH .

  THEREFORE K_C= [H₂O]¹ [C₂H₅CH₃COO]¹

                       ----------------------

                        [CH₃COOH ]¹ [C₂H₅OH]^{1      [     ]  SIGN DENOTES CONCENTRATION MEANS MOLES / VOLUME} 

                       K_C = [{1+(54.3*1)/100}/V] [{0+(54.3*1)/100}/V]

                        ------------------------------------------------

                         [{1-(54.3*1)/100}/V]  [{1-(54.3*1)/100}/V]   

SOLVE THIS EQUATION AND GET YOUR ANSWER

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