$n_{CH_{3}COOH}=\frac{7,2}{60}=0,12(mol)(1)$
$n_{C_{2}H_{5}OH}=\frac{4,6}{46}=0,1(mol)(2)$
Từ $(1)$,$(2)⇒$C_{2}H_{5}OH$ hết,$CH_{3}COOH$ dư
$PTHH: CH_{3}COOH+C_{2}H_{5}OH⇄CH_{3}COOC_{2}H_{5}+H_{2}O$
$mol:$ $0,1$ $\xrightarrow {H=80%}$ $0,08$
$m_{CH_{3}COOC_{2}H_{5}}=88.0,08=7,04(g)$