Em tham khảo nha :
\(\begin{array}{l}
a)\\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}\\
b)\\
{n_{{H_2}}} = \dfrac{{16,8}}{{22,4}} = 0,75mol\\
{n_{{C_2}{H_5}OH}} = 2{n_{{H_2}}} = 1,5mol\\
{m_{{C_2}{H_5}OH}} = 1,5 \times 46 = 69g\\
c)\\
C{H_3}COOH + {C_2}{H_5}OH \to C{H_3}COO{C_2}{H_5} + {H_2}O\\
{m_{C{H_3}COOH}} = \dfrac{{200 \times 60}}{{100}} = 120g\\
{n_{C{H_3}COOH}} = \dfrac{{120}}{{60}} = 2mol\\
\dfrac{2}{1} > \dfrac{{1,5}}{1} \Rightarrow C{H_3}COOH\text{ dư}\\
{n_{C{H_3}COO{C_2}{H_5}}} = {n_{{C_2}{H_5}OH}} = 1,5mol\\
{m_{C{H_3}COO{C_2}{H_5}}} = 1,5 \times 88 = 132g\\
\text{Khối lượng thu được sau phản ứng là :}\\
m = \dfrac{{132 \times 60}}{{100}} = 79,2g
\end{array}\)