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 `:`

Options

A. 44g
B. 20.33g
C. 22g
D. 58.66g

Answer

D

Explanation

Molar mass of `CH_(3)COOH=60g mol^(-1)`
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`.