Đáp án đúng: C
Giải chi tiết:\(\begin{gathered} {X_1}\left\{ \begin{gathered} Fe:0,1(BTNT:Fe) \hfill \\ FeC{O_3}:0,05(BTNT:\,C) \hfill \\ M:\frac{{2,7}}{M} \hfill \\ \end{gathered} \right.\xrightarrow{{HNO3du}}\left\langle \begin{gathered} {X_2}\left\{ \begin{gathered} Fe{(N{O_3})_3} \hfill \\ M{(N{O_3})_n} \hfill \\ N{H_4}N{O_3} \hfill \\ HN{O_3}du \hfill \\ \end{gathered} \right.\xrightarrow{{ + 0,2mol\,NaOH}}X3\left\{ \begin{gathered} Fe{(N{O_3})_3}:0,15 \hfill \\ M{(N{O_3})_n}:\frac{{2,7}}{M} \hfill \\ NaN{O_3}:0,2mol \hfill \\ N{H_4}N{O_3}:x \hfill \\ \end{gathered} \right. \hfill \\ Y1\left\{ \begin{gathered} CO2:0,05 \hfill \\ NO:0,15 \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} \right. \hfill \\ {X_3}\xrightarrow{{2P}}\left\{ \begin{gathered} {P_1}\xrightarrow{{co\,\,can}}38,3\,g \hfill \\ {P_2}\xrightarrow{{ + NaOH\,du}}8,025g\, \hfill \\ (Fe{(OH)_3} \downarrow :0,075\,mol) \hfill \\ \end{gathered} \right. \hfill \\ {m_{{X_3}}} = 38,3.2 = 76,6 = > 242.0,15 + \frac{{2,7}}{M}(M + 62n) + 85.0,2 + 80x = 76,6(1) \hfill \\ BT\,e:3{n_{Fe}} + {n_{FeC{O_3}}} + n.{n_M} = 3{n_{NO}} + 8{n_{N{H_4}N{O_3}}} = > 0,1.3 + 0,05 + \frac{{2,7n}}{M} = 3.0,15 + 8x(2) \hfill \\ (1)(2) = > \frac{n}{M} = \frac{1}{9};x = 0,025mol \hfill \\ = > n = 3;M = 27(Al);x = 0,025 \hfill \\ BTNT\,N:{n_{HN{O_3}}} = 3{n_{Fe{{(N{O_3})}_3}}} + 3{n_{Al{{(N{O_3})}_3}}} + {n_{NaN{O_3}}} + 2{n_{N{H_4}N{O_3}}} + {n_{NO}} \hfill \\ = 0,15.3 + 0,1.3 + 0,2 + 0,025.2 + 0,15 = 1,15mol \hfill \\ = > x = 1,15/0,5 = 2,3M \hfill \\ \end{gathered} \)
Đáp án C