Em tham khảo nha :
\(\begin{array}{l}
a)\\
2C{H_3}COOH + Mg \to {(C{H_3}COO)_2}Mg + {H_2}\\
b)\\
{m_{C{H_3}COOH}} = \dfrac{{150 \times 12}}{{100}} = 18g\\
{n_{C{H_3}COOH}} = \dfrac{{18}}{{60}} = 0,3mol\\
{n_{{H_2}}} = \dfrac{{{n_{C{H_3}COOH}}}}{2} = 0,15mol\\
{V_{{H_2}}} = 0,15 \times 22,4 = 3,36l\\
c)\\
{n_{Mg}} = \dfrac{{{n_{C{H_3}COOH}}}}{2} = 0,15mol\\
{m_{Mg}} = 0,15 \times 24 = 3,6g\\
d)\\
{n_{{{(C{H_3}COO)}_2}Mg}} = {n_{Mg}} = 0,15mol\\
{m_{{{(C{H_3}COO)}_{Mg}}}} = 0,15 \times 142 = 21,3g\\
{m_{ddspu}} = 150 + 3,6 - 0,15 \times 2 = 153,3g\\
C{\% _{{{(C{H_3}COO)}_2}Mg}} = \dfrac{{21,3}}{{153,3}} \times 100\% = 13,89\%
\end{array}\)
