Đáp án:
b) Mg: 7,2 g; Fe: 8,4 g
c) $C{\% _{{{(C{H_3}COO)}_2}Mg}} = 10,19\% ;C{\% _{{{(C{H_3}COO)}_2}Fe}} = 6,24\% $
Giải thích các bước giải:
a)
$\begin{gathered}
Mg + 2C{H_3}COOH \to {(C{H_3}COO)_2}Mg + {H_2} \hfill \\
Fe + 2C{H_3}COOH \to {(C{H_3}COO)_2}Fe + {H_2} \hfill \\
\end{gathered} $
b) ${n_{Mg}} = 2x;{n_{Fe}} = x$
Theo PTHH: ${n_{{H_2}}} = {n_{Mg}} + {n_{Fe}}$
$\begin{gathered}
\Leftrightarrow 0,45 = 2x + x = 3x \Rightarrow x = 0,15 \hfill \\
\Rightarrow {m_{Fe}} = 0,15.56 = 8,4g \hfill \\
{m_{Mg}} = 0,3.24 = 7,2g \hfill \\
\end{gathered} $
c) ${m_{ddC{H_3}COOH}} = D.V = 360.1,12 = 403,2g$
Bảo toàn khối lượng: ${m_X} + {m_{ddC{H_3}COOH}} = {m_{ddsau}} + {m_{{H_2}}}$
$\begin{gathered}
\Rightarrow {m_{ddsau}} = {m_X} + {m_{ddC{H_3}COOH}} - {m_{{H_2}}} \hfill \\
= 7,2 + 8,4 + 403,2 - 0,45.2 = 417,9g \hfill \\
\end{gathered} $
$\begin{gathered}
{n_{{{(C{H_3}COO)}_2}Mg}} = {n_{Mg}} = 0,3mol \Rightarrow {m_{{{(C{H_3}COO)}_2}Mg}} = 42,6g \hfill \\
{n_{{{(C{H_3}COO)}_2}Fe}} = {n_{Fe}} = 0,15mol \Rightarrow {m_{{{(C{H_3}COO)}_2}Fe}} = 26,1g \hfill \\
\Rightarrow C{\% _{{{(C{H_3}COO)}_2}Mg}} = \dfrac{{42,6}}{{417,9}}.100\% = 10,19\% \hfill \\
C{\% _{{{(C{H_3}COO)}_2}Fe}} = \dfrac{{26,1}}{{417,9}}.100\% = 6,24\% \hfill \\
\end{gathered} $
