Đáp án đúng: A
Giải chi tiết:nFe = 1,12 : 56 = 0,02 (mol) ; nHCl = 0,3.0,2 = 0,06(mol);
\(\begin{gathered} \underbrace {Fe}_{0,02\,mol} + \underbrace {HCl}_{0,06\,mol}\xrightarrow[{}]{}\left\langle \begin{gathered} {H_2}:0,02\,mol \hfill \\ \left\{ \begin{gathered} FeCl{}_2:0,02\,mol \hfill \\ HCl\,du\,:0,02\,mol \hfill \\ \end{gathered} \right.\xrightarrow{{ + AgN{O_3}\,du}}\left\{ \begin{gathered} A{g^ + }\,\,\, + \,\,\,C{l^ - }\xrightarrow{{}}AgCl \downarrow \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,0,06\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,0,06 \hfill \\ 3F{e^{2 + }} + 4{H^ + } + N{O_3}^ - \xrightarrow{{}}3F{e^{3 + }} + NO + 2{H_2}O \hfill \\ 0,015 \leftarrow 0,02 \hfill \\ F{e^{2 + }}\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,\,A{g^ + }\xrightarrow{{}}F{e^{3 + }} + Ag \downarrow \hfill \\ (0,02 - 0,015)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,0,005\,\,\, \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} \right. \hfill \\ m \downarrow = {m_{Ag}}\, + \,{m_{AgCl}} = 0,005.108 + 0,06.143,5 = 9,15(g) \hfill \\ \end{gathered} \)
Đáp án A
