Em tham khảo nha :

\(\begin{array}{l}
a)\\
{C_2}{H_5}OH + 3{O_2} \to 2C{O_2} + 3{H_2}O\\
C{H_3}COOH + 2{O_2} \to 2C{O_2} + 2{H_2}O\\
C{H_3}COOH + NaOH \to C{H_3}COONa + {H_2}O\\
{n_{C{O_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
{n_{NaOH}} = 0,03 \times 1 = 0,03mol\\
{n_{C{H_3}COOH}} = {n_{NaOH}} = 0,03mol\\
{n_{{C_2}{H_5}OH}} = \dfrac{{{n_{C{O_2}}} - 2{n_{C{H_3}COOH}}}}{2} = 0,07mol\\
{m_{C{H_3}COOH}} = 0,03 \times 60 = 1,8g\\
{m_{{C_2}{H_5}OH}} = 0,07 \times 46 = 3,22g\\
\% C{H_3}COOH = \dfrac{{1,8}}{{1,8 + 3,22}} \times 100\%  = 35,9\% \\
\% {C_2}{H_5}OH = 100 - 35,9 = 64,1\% \\
b)\\
C{H_3}COOH + {C_2}{H_5}OH \to C{H_3}COO{C_2}{H_5} + {H_2}O\\
\dfrac{{0,03}}{1} < \dfrac{{0,07}}{1} \Rightarrow {C_2}{H_5}OH\text{ dư}\\
{n_{C{H_3}COO{C_2}{H_5}}} = {n_{C{H_3}COOH}} = 0,03mol\\
{m_{C{H_3}COO{C_2}{H_5}}} = 0,03 \times 88 = 2,64g\\
\text{Khối lượng este thu được là :}\\
m = \dfrac{{2,64 \times 30}}{{100}} = 0,792g
\end{array}\)