Đáp án:
\({m_m} = 34,165g\)
Giải thích các bước giải:
\(\begin{array}{l}
Mg + 2HCl \to MgC{l_2} + {H_2}(1)\\
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O(2)\\
A{l_2}{O_3} + 6HCl \to 2AlC{l_3} + 3{H_2}O(3)\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
{n_{MgC{l_2}}} = {n_{{H_2}}} = 0,1mol\\
{m_{MgC{l_2}}} = n \times M = 0,1 \times 95 = 9,5g\\
{n_{HCl(1)}} = 2{n_{{H_2}}} = 0,2mol\\
{n_{HCl}} = {C_M} \times V = 0,1 \times 7 = 0,7mol\\
hh:F{e_2}{O_3}(a\,mol),A{l_2}{O_3}(b\,mol)\\
\left\{ \begin{array}{l}
a - b = 0\\
6a + 6b = 0,5
\end{array} \right.\\
\Rightarrow a = \dfrac{1}{{24}}mol;b = \dfrac{1}{{24}}mol\\
{n_{FeC{l_3}}} = 2{n_{F{e_2}{O_3}}} = \dfrac{1}{{12}}mol\\
{m_{FeC{l_3}}} = n \times M = \dfrac{1}{{12}} \times 162,5 = 13,54g\\
{n_{AlC{l_3}}} = 2{n_{A{l_2}{O_3}}} = \dfrac{1}{{12}}mol\\
{m_{AlC{l_3}}} = n \times M = \dfrac{1}{{12}} \times 133,5 = 11,125g\\
{m_m} = {m_{MgC{l_2}}} + {m_{FeC{l_3}}} + {m_{AlC{l_3}}} = 9,5 + 13,54 + 11,125 = 34,165g
\end{array}\)
