On taking 60 g `CH_(3)`COOH and 46g `CH_(3)CH_(2)`OH in a 5 lit. lask in tle presence of `H_(3)O^(+)` (catalyst), at 298K 44 g of `CH_(3)COOC_(2)H_(5)` is formed at equilibrium. If amount of `CH_(3)`COOH is doubled without affecting amount of `CH_(3)CH_(2)OH` then, `CH_(3)"COOC"_(2)H_(5)` formed is

Options

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

Answer

D

Explanation

Case-I
`CH_(3)COOH=(60)/(60)=1` mol
`CH_(3)CH_(2)OH=(46)/(46)=1` mol
`CH_(3)COOC_(2)H_(5)=(44)/(88)=0.5` mol
Case-II
`CH_(3)COOH=2 mol, C_(2)H_(5)OH=1` mol `CH_(3)COOO_(2)H_(5)` =?

`K_(c)` is not affected at temperature is constant `(x^(2))/(1-x^(2))=(y^(2))/((2-y)(1-y)), (y^(2))/((2-y)(1-y))=1` `y^(2)=(2-4)(1-4)=y^(2)-3y`
`y=(2)/(3)` mole of ester = `(2)/(3) xx 58 = 58.66` g