1.

The equilibrium constant for the reaction `CH_(3) COOH + C_(2)H_(5)OH hArrCH_(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 alchol.

Answer» Correct Answer - `117*04 g`
Initially, `CH_(3) COOH = 120/60 "mol" = 2 "mol"`
`C_(2) H_(5) OH= 92/46 "mol" = 2 "mol"`
At equilibrium, `[CH_(3)COOH] = (2-x)//V " mol "L^(-1)`
`[C_(2)H_(5)OH]= (2-x)//V " mol "L^(-1),`
`[CH_(3)COOC_(2)H_(5)]=[H_(2)O]= x//V " mol " L^(-1)`
` K = (x xx x )/(2-x)^(2) = 4 " (Given)"`.
`" This gives "x== 1*33 " mol"`
` :. " Mass of ethyl acetate " = 1*33 xx 88 = 117*04 g " "("Molar mass of " CH_(3)COOC_(2)H_(5)= 88" g mol"^(-1))`


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