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