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The equilibrium constant for the reacti...

The equilibrium constant for the reaction :
`CH_(3) COOH + C_(2)H_(5)OH hArr CH_(3) COOC_(2)H_(5) +H_(2)O`
is 4.0 at `25^(@)C`. Calculate the weight of ethyl acetate that will be obtained when 120 g of acetic acid are reacted with 92 g of ethyl alcohol.

Text Solution

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The given reaction is :
`CH_(3)COOH + C_(2)H_(5)OH hArr CH_(3) COOC_(2) H_(5) +H_(2) O`
Applying the Law of Chemical equilibrium,
`K_(c) = [[CH_(3)COOC_(2)H_(5)][H_(2)O]]/[[CH_(3)COOH][C_(2)H_(5)OH]]=4.0`
Initial moles of `CH_(3)COOH = (120g)/(60 g mol^(-1)) = 2` mol (Molar mass of `CH_(3) COOH = 60 g mol^(-1))`
Initial moles of `C_(2) H_(5) OH = (92g)/(46 mol^(-1)) =2` mol (Molar mass of `C_(2)H_(5)OH= 46 g mol^(-1))`
Let the moles of `CH_(3) COOH` and `C_(2)H_(5)OH` reacted be x. Therefore , according to the equation x,moles of `CH_(3)COOC_(2)H_(5)" and "H_(2)O` will be formed .
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