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60g CH_(3)COOH and 46g C_(2)H_(5)OH react in 5L flask to form 44g CH_(3)COOC_(2)H_(5) at equilibrium. On taking 120 g CH_(3)COOH and 46g C_(2)H_(5)OH, CH_(3)COOC_(2)H_(5) formed at equilibrium is : |
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Answer» 44g Molar mass of `C_(2)H_(5)OH=46g mol^(-1)` Molar mass of `CH_(3)COOC_(2)H_(5)=88g mol^(-1)` `:. [CH_(3)COOH)_("Initial")=(60)/(60xx5)=0.2 "mol"L^(-1)` `[C_(2)H_(5)OH]_("Initial")=(46)/(46xx5)=0.2"mol" L^(-1)` `[CH_(3)COOC_(2)H_(5)]_(eqm)=(44)/(88)xx(1)/(5)=0.1 "mol" L^(-1)` `CH_(3)COOH+C_(2)H_(5)OH hArr CH_(3)COOC_(2)H_(5)+H_(2)O` Initial `0.2 M ""0.2M` At eqm. `(0.2-0.1)M""(0.2-0.1)M""0.1M""0.1M` `:. K=([CH_(3)COOC_(2)H_(5)][H_(2)O])/([CH_(3)COOH][C_(2)H_(5)OH])` `=(0.1xx0.1)/(0.1xx0.1)=1` In second case, `[CH_(3)COOH]_("Initial")=0.4M` `[C_(2)H_(5)OH]_("Initial")=0.2M` If x is the amount of acid and ALCOHOL reacted `[CH_(3)COOH]_("eqm.")=(0.40-x)M` `[C_(2)H_(5)OH]_("eqm.")=[0.2-x]M` `[CH_(3)COOC_(2)H_(5)]_("eqm.")=[H_(2)O]_("eqm.")=xM` `:. K=(x^(2))/((0.4-x)(0.2-x))=1` `:. x^(2)=(0.4-x)(0.2-x)` `x^(2)=0.08+x^(2)-0.6x` `0.6x=0.08` `x=(8)/(60)M` `:. `Moles of ethyl acetate produced `=(8)/(60)XX5=(2)/(3)` Mass of ethyl acetate produced `=(2)/(3)xx88=58=58.66g`. |
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