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