\(n_{CH_3COOH}=\dfrac{9}{60}=0,15\left(mol\right)\\ n_{C_2H_5OH}=\dfrac{8,28}{46}=0,18\left(mol\right)\\ n_{CH_3COOC_2H_5}=\dfrac{10,56}{88}=0,12\left(mol\right)\)
PTHH: \(C_2H_5OH+CH_3COOH\xrightarrow[t^o]{H_2SO_4đặc}CH_3COOC_2H_5+H_2O\)
0,12 <------- 0,12 <------------------------ 0,12
LTL: 0,18 > 0,15 => C2H5OH dư, tính theo CH3COOH
\(H=\dfrac{0,12}{0,15}=80\%\\ \rightarrow B\)
