Đáp án:
$B$
Giải thích các bước giải:
$\begin{gathered} Fe + 2HCl \to FeC{l_2} + {H_2}\,\,\,\,\,\,\, \hfill \\ FeO + 2HCl \to FeC{l_2} + {H_2}O \hfill \\ \end{gathered} $
Theo PTHH: ${n_{{H_2}}} = {n_{Fe}} = 0,2\,\,mol$
$\begin{gathered}
\to {m_{FeO}} = 16,96 - 0,2.56 = 5,76\,\,gam \hfill \\
\to {n_{FeO}} = 0,08\,\,mol\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \hfill \\
\end{gathered} $
$\to {n_{FeC{l_2}}} = 0,08 + 0,2 = 0,28\,\,mol$
Dung dịch X chứa $FeCl_2 (0,28 mol), HCl (0,76 - 0,28.2=0,2 mol)$ dư
Thêm $AgNO_3$ vào dung dịch X:
PTHH:
$A{g^ + } + C{l^ - } \to AgCl \downarrow $
${n_{AgCl}} = {n_{C{l^ - }}} = 0,76\,\,mol$
$\begin{gathered} \,\,\,\,\,\,\,\,\,\,\,\,\,4{H^ + } + NO_3^ - + 3F{e^{2 + }} \to 3F{e^{3 + }} + NO + 2{H_2}O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \hfill \\ 0,2 \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,15\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol \hfill \\ F{e^{2 + }} + A{g^ + } \to F{e^{3 + }} + Ag\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \hfill \\ 0,13 \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,13\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol \hfill \\ \end{gathered} $
Kết tủa: $AgCl (0,76 mol); Ag (0,13 mol)$
$ \to m = 0,76.143,5 + 0,13.108 = 123,1\,\,gam$
Chọn B
