\(n_{C_2H_5OH}=\dfrac{138}{46}=3\left(mol\right)\)
\(n_{CH_3COOH}=\dfrac{180}{60}=3\left(mol\right)\)
PT: CH3COOH + C2H5OH \(\underrightarrow{Axit}\) CH3COOC2H5 + H2O
\(n_{CH_3COOH}\) phản ứng = \(3\cdot44\%=1.32\left(mol\right)\)
\(\Rightarrow n_{CH_3COOC_2H_5}=1.32\left(mol\right)\)
\(\Rightarrow m_{CH_3COOC_2H_5}=1.32\cdot88=116.16\left(gam\right)\)