95%乙醇 6ml大约是0.1摩尔酒精
CH3COOH+C2H5OH=CH3COOC2H5+H2O
设生成CH3COOC2H5Xmol,H2OXmol,则有X^2/(0.1-X)*(0.09-X)=3.77,得X=1/16,所以乙酸乙酯实验产率为1/16/0.09=70%
95%乙醇 6ml大约是0.1摩尔酒精
CH3COOH+C2H5OH=CH3COOC2H5+H2O
设生成CH3COOC2H5Xmol,H2OXmol,则有X^2/(0.1-X)*(0.09-X)=3.77,得X=1/16,所以乙酸乙酯实验产率为1/16/0.09=70%