Question

Answer 1

nCH3COOH=12/60=0,2(mol)

nCH3COOC2H5(TT)=11/88=0,125(mol)

PTHH:CH3COOH + C2H5OH \(⇌\) (H⁺ , t^o)   CH3COOC2H5

nCH3COOC2H5(LT)=nCH3COOH=0,2(mol)

=>H= (0,125/0,2).100=62,5%

Answer 2

Bài 9: nCH3COOH= 6/60=0,1(mol)

nC2H5OH= (6/46)= 3/23(mol)

PTHH:CH3COOH + C2H5OH \(⇌\) (H⁺ , t^o)   CH3COOC2H5

Ta có: 0,1/1 < 3/23 :1

=> CH3COOH hết, C2H5OH dư

Ta có: nCH3COOC2H5(LT)= nCH3COOH=0,1(mol)

Vì: H=80% => nCH3COOC2H5(TT)= 80%. 0,1=0,08(mol)

=>mCH3COOC2H5(TT)= 0,08.88=7,04(g)