Em tham khảo nha :
\(\begin{array}{l}
a)\\
2C{H_3}COOH + Zn \to {(C{H_3}COO)_2}Zn + {H_2}\\
{m_{C{H_3}COOH}} = \dfrac{{240 \times 10}}{{100}} = 24g\\
{n_{C{H_3}COOH}} = \dfrac{{24}}{{60}} = 0,4mol\\
{n_{{H_2}}} = \dfrac{{{n_{C{H_3}COOH}}}}{2} = 0,2mol\\
{V_{{H_2}}} = 0,2 \times 22,4 = 4,48l\\
b)\\
{n_{Zn}} = \dfrac{{{n_{C{H_3}COOH}}}}{2} = 0,2mol\\
{m_{Zn}} = 0,2 \times 65 = 13g\\
c)\\
{m_{{\rm{dd}}spu}} = 13 + 240 - 0,2 \times 2 = 252,6g\\
{n_{{{(C{H_3}COO)}_2}Zn}} = {n_{Zn}} = 0,2mol\\
{m_{{{(C{H_3}COO)}_2}Zn}} = 0,2 \times 183 = 36,6g\\
C\% = \dfrac{{36,6}}{{252,6}} \times 100\% = 14,5\% \\
d)\\
C{H_3}COOH + {C_2}{H_5}OH \to C{H_3}COO{C_2}{H_5} + {H_2}O\\
{n_{C{H_3}COO{C_2}{H_5}}} = {n_{C{H_3}COOH}} = 0,4mol\\
{m_{C{H_3}COO{C_2}{H_5}}} = 0,4 \times 88 = 35,2g\\
H = \dfrac{{8,8}}{{35,2}} \times 100\% = 25\%
\end{array}\)