Question: TYPE 3: HARDER STILL (Using initial concentrations where volume does not cancel) Ethyl ethanoate (CH,CO,C Hs) can be formed by the reaction of ethene (C2H.)

 TYPE 3: HARDER STILL (Using initial concentrations where volume does not

TYPE 3: HARDER STILL (Using initial concentrations where volume does not cancel) Ethyl ethanoate (CH,CO,C Hs) can be formed by the reaction of ethene (C2H.) with ethanoic acid (CH.COOH) in an inert solvent according to the equation C2H2 + CH3COOH - CH,CO,C Hs In an experiment 0.5 moles of ethene was allowed to react with 0.2 moles of ethanoic acid at 10C, the total volume being made up to 250cm with an inert solvent. When equilibrium had been established the mixture was found to contain 0.18 moles of ethyl ethanoate. Calculate the number of moles of ethene and ethanoic acid present at equilibrium, and hence the molar concentration of each substance present at equilibrium and hence the value of Kc for the reaction under these conditions

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