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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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