Em tham khảo nha :
\(\begin{array}{l}
a)\\
C{H_3}COOH + {C_2}{H_5}OH \to C{H_3}COO{C_2}{H_5} + {H_2}O\\
b)\\
{n_{C{H_3}COOH}} = \dfrac{{12}}{{60}} = 0,2mol\\
{n_{{C_2}{H_5}OH}} = \dfrac{{13,8}}{{46}} = 0,3mol\\
\dfrac{{0,2}}{1} < \dfrac{{0,3}}{1} \Rightarrow {C_2}{H_5}OH\text{ dư}\\
{n_{{C_2}{H_5}O{H_d}}} = 0,3 - 0,2 = 0,1mol\\
{m_{{C_2}{H_5}O{H_d}}} = 0,1 \times 46 = 4,6g\\
c)\\
{n_{C{H_3}COO{C_2}{H_5}}} = {n_{C{H_3}COOH}} = 0,2mol\\
{m_{C{H_3}COO{C_2}{H_5}}} = 0,2 \times 88 = 17,6g\\
H = \dfrac{{14,08}}{{17,6}} \times 100\% = 80\% \\
d)\\
{V_{{C_2}{H_5}OH}} = \dfrac{{13,8}}{{0,8}} = 17,25ml\\
D = \dfrac{{17,25}}{{17,25 + 40,25}} \times 100 = {30^0}
\end{array}\)