Câu 1:

2CH3COOH+2Na−−−>2CH3COONa+H22CH3COOH+2Na−−−>2CH3COONa+H2(1)(1)

2C2H5OH+2Na−−−>2C2H5ONa+H22C2H5OH+2Na−−−>2C2H5ONa+H2(2)(2)

Gọi a, b lần lượt là số mol của CH3COOH,C2H5OHCH3COOH,C2H5OHcó trong hỗn hợp

nH2(đktc)=0,33622,4=0,015(mol)nH2(đktc)=0,33622,4=0,015(mol)

Theo PTHH (1) và (2) nH2=0,5a+0,5b(mol)nH2=0,5a+0,5b(mol)

=>0,5a+0,5b=0,015=>0,5a+0,5b=0,015 (I)(I)

CH3COOH+NaOH−−−>CH3COONa+H2OCH3COOH+NaOH−−−>CH3COONa+H2O(3)(3)

nNaOH=0,1.0,2=0,02(mol)nNaOH=0,1.0,2=0,02(mol)

Theo PTHH (3) nCH3COOH=nNaOH=0,02(mol)nCH3COOH=nNaOH=0,02(mol)

=>a=0,02=>a=0,02 (II)(II)

Giai he (I) và (II) ⇒{a=0,02b=0,01⇒{a=0,02b=0,01

mCH3COOH=0,02.60=1,2(g)mCH3COOH=0,02.60=1,2(g)

mC2H5OH=0,01.46=0,46(g)mC2H5OH=0,01.46=0,46(g)

=>m=1,2+0,46=1,66(g)=>m=1,2+0,46=1,66(g)

%mCH3COOH=1,2.1001,66=72,29%%mCH3COOH=1,2.1001,66=72,29%

=>%mC2H5OH=100%−72,29%=27,71%