a,
$CH_{3}COOH+C_{2}H_{5}OH \to CH_{3}COOC_{2}H_{5}+H_{2}O \\\text{0,1 0,05
ban đầu}\\\text{0,05 0,05
0,05 phản ứng}\\\text{0,05 0 0,05 sau phản ứng} $
$nCH_{3}COOH=\frac{6}{60}=0,1$
$nC_{2}H_{5}OH=\frac{2,3}{46}=0,05$
$nCH_{3}COOC_{2}H_{5}=\frac{3,52}{88}=0,04$
$H=\frac{0,04}{0,05}.100=80\%$
b,
$nC_{2}H_{5}OHdư=0,05.(1-80\%)=0,01$
$mC_{2}H_{5}OHdư=0,01.46=0,46g$
$VC_{2}H_{5}OHdư=\frac{0,46}{0,8}=0,575ml$
$VH_{2}O=\frac{20,7}{1}=20,7ml$
$\text{Độ rượu}=\frac{0,575}{20,7+0,575}.100=2,7^o$