Đáp án:
Giải thích các bước giải:
a)
$KOH+CH3COOH-->CH3COOK+H2O$
$nKOH=0,2.0,4=0,08(mol)$
theo PTHH
$nCH3COOH=nKOH=0,08(mol)$
$=>mCH3COOH=0,08.46=3,68(g)$
b)
$CH3COOH+C2H5OH---H2SO4đ,to->CH3COOC2H5+H2O$
$nCH3COOC2H5=nCH3COOH=0,08(mol)$
$=>mCH3COOC2H5=0,08.88=7,04(g)$
Do H=70%
=>$mCH3COOC2H5=7,04.70$%$=4,928(g)$