Đáp án:
a) Sắt
b) 72,06% và 27,94%
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
M + 2HCl \to MC{l_2} + {H_2}\\
2MC{l_2} + C{l_2} \to 2MC{l_3}\\
MC{l_2} + 2NaOH \to M{(OH)_2} + 2NaCl\\
MC{l_3} + 3NaOH \to M{(OH)_3} + 3NaCl\\
{n_{M{{(OH)}_2}}} = \dfrac{{19,8}}{{{M_M} + 34}}\,\,mol\\
{n_{MC{l_2}}} = {n_{M{{(OH)}_2}}} = \dfrac{{19,8}}{{{M_M} + 34}}\,mol\\
{m_{MC{l_2}}} = \dfrac{{19,8}}{{{M_M} + 34}} \times (M + 71) = \dfrac{{19,8M + 1405,8}}{{{M_M} + 34}}g\\
{m_{MC{l_2}}} = 0,5{M_M} \Leftrightarrow \frac{{19,8M + 1405,8}}{{{M_M} + 34}} = 0,5{M_M}\\
\Leftrightarrow 0,5{M_M}^2 + 17{M_M} = 19,8{M_M} + 1405,8\\
\Leftrightarrow 0,5{M_M}^2 - 2,8{M_M} - 1405,8 = 0 \Leftrightarrow \left[ \begin{array}{l}
{M_M} \approx 56\\
{M_M} \approx - 50
\end{array} \right.\\
\Rightarrow M:Fe\\
b)\\
{n_{FeC{l_2}}} = {n_{Fe{{(OH)}_2}}} = \dfrac{{19,8}}{{90}} = 0,22\,mol\\
\% {m_{FeC{l_2}}} = \dfrac{{0,22 \times 127}}{{100}} \times 100\% = 27,94\% \\
\% {m_{FeC{l_3}}} = 100 - 27,94 = 72,06\%
\end{array}\)
