Đáp án:
\(\begin{array}{l}
b){V_{{H_2}}} = 4,48l\\
c)C{\% _{Mg{{(C{H_3}{\rm{COO}})}_2}}} = 42,77\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)2C{H_3}{\rm{COO}}H + Mg \to Mg{(C{H_3}{\rm{COO}})_2} + {H_2}\\
b)\\
{n_{Mg}} = 0,2mol\\
\to {n_{{H_2}}} = {n_{Mg}} = 0,2mol\\
\to {V_{{H_2}}} = 4,48l
\end{array}\)
\(\begin{array}{l}
c)\\
{n_{Mg{{(C{H_3}{\rm{COO}})}_2}}} = {n_{Mg}} = 0,2mol\\
\to {m_{Mg{{(C{H_3}{\rm{COO}})}_2}}} = 28,4g\\
{m_{{\rm{dd}}}} = {m_{Mg}} + {m_{{\rm{dd}}C{H_3}{\rm{COO}}H}} - {m_{{H_2}}} = 66,4\\
\to C{\% _{Mg{{(C{H_3}{\rm{COO}})}_2}}} = \dfrac{{28,4}}{{66,4}} \times 100\% = 42,77\%
\end{array}\)
