a) PTHH: C2H5OH + CH3COOH \(\underrightarrow{H_2SO_4đ,t^o}\) CH3COOC2H5 + H2O
b) \(n_{CH_3COOH}=\dfrac{60}{60}=1\left(mol\right)\\ n_{C_2H_5OH}=\dfrac{36,8}{46}=0,8\left(mol\right)\)
So sánh tỉ lệ, ta thấy:
\(\dfrac{1}{1}>\dfrac{0,8}{1}\)
=> CH3COOH dư, C2H5OH hết. Tính theo \(n_{C_2H_5OH}\).
Ta có:\(n_{CH_3COOC_2H_5\left(TT\right)}=\dfrac{42,24}{88}=0,48\left(mol\right)\)
\(n_{CH_3COOC_2H_5\left(LT\right)}=n_{C_2H_5OH}=0,8\left(mol\right)\)
=> \(H=\dfrac{0,48}{0,8}.100=60\%\)