Em tham khảo nha:
\(\begin{array}{l}
2)\\
{C_2}{H_4} + {H_2}O \xrightarrow{xt,t^0} {C_2}{H_5}OH\\
{C_2}{H_5}OH + {O_2} \xrightarrow{\text{ lên men }} C{H_3}COOH + {H_2}O\\
C{H_3}COOH + {C_2}{H_5}OH\xrightarrow{xt,t^0} C{H_3}COO{C_2}{H_5} + {H_2}O\\
3)\\
a)\\
C{H_3}COOH + NaOH \to C{H_3}COONa + {H_2}O\\
{n_{NaOH}} = 0,05 \times 1 = 0,05\,mol\\
{n_{C{H_3}COOH}} = {n_{NaOH}} = 0,05\,mol\\
{m_{C{H_3}COOH}} = 0,05 \times 60 = 3g\\
{m_{{C_2}{H_5}OH}} = 7,6 - 3 = 4,6g\\
b)\\
{n_{{C_2}{H_5}OH}} = \dfrac{{4,6}}{{46}} = 0,1\,mol\\
C{H_3}COOH + {C_2}{H_5}OH \xrightarrow{xt,t^0} C{H_3}COO{C_2}{H_5} + {H_2}O\\
{n_{C{H_3}COOH}} < {n_{{C_2}{H_5}OH}} \Rightarrow \text{$C_2H_5OH$ dư tính theo $CH_3COOH$ }\\
{n_{C{H_3}COO{C_2}{H_5}}} = {n_{C{H_3}COOH}} = 0,05\,mol\\
{m_{C{H_3}COO{C_2}{H_5}}} = 0,05 \times 88 = 4,4g
\end{array}\)
