Đáp án:

b.

$\% {m_{Mg}} = 42,48\% $
$\% {m_{Zn}} = 57,52\% $

c.

$C{\% _{C{H_3}{\rm{COO}}H}} = 18\%$

$C{\% _{Mg{{(C{H_3}{\rm{COO)}}}_2}}}  = 13,48\% $
$ C{\% _{Zn{{(C{H_3}{\rm{COO)}}}_2}}}  = 8,69\% $

Giải thích các bước giải:

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

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

\(\begin{array}{l}
\left\{ \begin{array}{l}
24a + 65b = 11,3\\
a + b = 0,3
\end{array} \right. \to \left\{ \begin{array}{l}
a = 0,2\\
b = 0,1
\end{array} \right.\\
 \to {m_{Mg}} = 0,2 \times 24 = 4,8g\\
 \to {m_{Zn}} = 0,1 \times 65 = 6,5g\\
b)\\
\% {m_{Mg}} = \dfrac{{4,8}}{{11,3}} \times 100\%  = 42,48\% \\
\% {m_{Zn}} = \dfrac{{6,5}}{{11,3}} \times 100\%  = 57,52\% \\
c)\\
{n_{C{H_3}{\rm{COO}}H}} = 2{n_{Mg}} + 2{n_{Zn}} = 0,6mol\\
 \to {m_{C{H_3}{\rm{COO}}H}} = 0,6 \times 60 = 36g\\
 \to C{\% _{C{H_3}{\rm{COO}}H}} = \dfrac{{36}}{{200}} \times 100\%  = 18\% \\
 \to {n_{Mg{{(C{H_3}{\rm{COO)}}}_2}}} = {n_{Mg}} = 0,2mol\\
 \to {n_{Zn{{(C{H_3}{\rm{COO)}}}_2}}} = {n_{Zn}} = 0,1mol\\
 \to {m_{Mg{{(C{H_3}{\rm{COO)}}}_2}}} = 0,2 \times 142 = 28,4g\\
 \to {m_{Zn{{(C{H_3}{\rm{COO)}}}_2}}} = 0,1 \times 183 = 18,3g\\
{n_{{H_2}}} = {n_{Mg}} + {n_{Zn}} = 0,3mol \to {m_{{H_2}}} = 0,6g\\
 \to {m_{{\rm{dd}}muối}} = {m_{hỗnhợp}} + {m_{{\rm{dd}}C{H_3}{\rm{COO}}H}} - {m_{{H_2}}}\\
 \to {m_{{\rm{dd}}muối}} = 11,3 + 200 - 0,6 = 210,7g\\
 \to C{\% _{Mg{{(C{H_3}{\rm{COO)}}}_2}}} = \dfrac{{28,4}}{{210,7}} \times 100\%  = 13,48\% \\
 \to C{\% _{Zn{{(C{H_3}{\rm{COO)}}}_2}}} = \dfrac{{18,3}}{{210,7}} \times 100\%  = 8,69\% 
\end{array}\)