Đáp án:

\(\begin{array}{l}
b,\\
\% {m_{Mg}} = 42,48\% \\
\% {m_{Zn}} = 57,52\% \\
c,\\
C{\% _{C{H_3}{\rm{COOH}}}} = 18\% \\
C{\% _{Mg{{(C{H_3}{\rm{COO}})}_2}}} = 13,48\% \\
C{\% _{{{(C{H_3}{\rm{COO}})}_2}Zn}} = 8,69\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a,\\
Mg + 2C{H_3}{\rm{COOH}} \to Mg{(C{H_3}{\rm{COO}})_2} + {H_2}\\
Zn + 2C{H_3}{\rm{COOH}} \to {(C{H_3}{\rm{COO}})_2}Zn + {H_2}\\
b,\\
{n_{{H_2}}} = 0,3mol
\end{array}\)

Gọi a và b lần lượt là số mol của Mg và Zn

\(\begin{array}{l}
\left\{ \begin{array}{l}
a + b = 0,3\\
24a + 65b = 11,3
\end{array} \right. \to \left\{ \begin{array}{l}
a = 0,2\\
b = 0,1
\end{array} \right.\\
 \to {n_{Mg}} = 0,2mol\\
 \to {n_{Zn}} = 0,1mol\\
 \to \% {m_{Mg}} = \dfrac{{0,2 \times 24}}{{11,3}} \times 100\%  = 42,48\% \\
 \to \% {m_{Zn}} = 100\%  - 42,48\%  = 57,52\% 
\end{array}\)

\(\begin{array}{l}
c,\\
{n_{C{H_3}{\rm{COOH}}}} = 2{n_{Mg}} + 2{n_{Zn}} = 0,6mol\\
 \to {m_{C{H_3}{\rm{COOH}}}} = 36g\\
 \to C{\% _{C{H_3}{\rm{COOH}}}} = \dfrac{{36}}{{200}} \times 100\%  = 18\% \\
{n_{Mg{{(C{H_3}{\rm{COO}})}_2}}} = {n_{Mg}} = 0,2mol\\
 \to {m_{Mg{{(C{H_3}{\rm{COO}})}_2}}} = 28,4g\\
{n_{{{(C{H_3}{\rm{COO}})}_2}Zn}} = {n_{Zn}} = 0,1mol\\
 \to {m_{{{(C{H_3}{\rm{COO}})}_2}Zn}} = 18,3g\\
{m_{{\rm{dd}}}} = {m_{hh}} + {m_{{\rm{dd}}}}_{C{H_3}{\rm{COOH}}} - {m_{{H_2}}} = 210,7g\\
 \to C{\% _{Mg{{(C{H_3}{\rm{COO}})}_2}}} = \dfrac{{28,4}}{{210,7}} \times 100\%  = 13,48\% \\
 \to C{\% _{{{(C{H_3}{\rm{COO}})}_2}Zn}} = \dfrac{{18,3}}{{210,7}} \times 100\%  = 8,69\% 
\end{array}\)