Hoà tan hoàn toàn 11,2648 gam hỗn hợp X gồm Fe, Fe(NO3)2 và Al vào dung dịch HCl

11,2648 (g) \[\left\{ \begin{array}{l} Fe\\ Fe{(N{O_3})_2}\\ Al \end{array} \right.\, + HC{l_{(Vua\,du)}} \to \,\,\left\{ \begin{array}{l} d.d\,Y\,\left\{ \begin{array}{l} F{e^{2 + }}\\ F{e^{3 + }}\\ NH_4^ + :\,a\,mol\\ A{l^{3 + }}\\ C{l^ - } \end{array} \right.\, + \,AgN{O_{3\,du}} \to \,72,2(g)\, \downarrow \left\{ \begin{array}{l} AgCl\\ Ag \end{array} \right.\\ \uparrow \left\{ \begin{array}{l} {N_2}:\,0,02\,mol\\ {H_2}:\,0,01\,mol \end{array} \right.\, + \,{H_2}O \end{array} \right.\,\,\] $\begin{array}{l} 12{H^ + }\,\, + \,2NO_3^ - \,\, + 10e\, \to \,{N_2}\, + \,{H_2}O\,\\ 0,24..............................0,02....0,12 \end{array}$ $\begin{array}{l} 10{H^ + }\,\, + \,NO_3^ - \, + 8e\, \to \,NH_4^ + \,\, + \,3{H_2}O\\ 10a.............................a...........3a \end{array}$ $\begin{array}{l} 2{H^ + }\,\, + \,2e\,\, \to \,{H_2}\, \uparrow \\ 0,02..............0,01 \end{array}$ → ${n_{HCl}}\, = \,('0,26 + 10a)\,mol\, = \,{n_{C{l^ - }\,\,}} = \,{n_{AgCl}}$ → ${n_{{H_2}O}}\, = \,0,12 + 3a\,(mol)$ Bảo toàn khối lượng: 11,2468 + 36,5 (0,26 +10a) = 24,2348 + 0,02.28 + 0,01. 2+ 18. (0,12+3a) → a = 0,02 → ${n_{C{l^ - }}} = \,0,26\, + 10a = \,0,46\,mol\, = \,{n_{AgCl}}$ ${n_{Ag}} = \,\frac{{72,2092 - 0,46.143,9}}{{108}} = 0,0574\,mol$ $F{e^{2 + }}\, + A{g^ + }\,\, \to \,Ag\, + \,F{e^{3 + }}$ → ${n_{F{e^{2 + }}}}\, = \,{n_{Ag}}\, = \,0,574\,mol$ Bảo toàn điện tích dung dịch Y: $2{n_{F{e^{2 + }}\,}} + \,3{n_{F{e^{3 + }}}}\,\, + \,{n_{NH_4^ + }}\, + \,3{n_{A{l^{3 + }}}}\, = \,{n_{C{l^ - }}}$ → $3{n_{F{e^{3 + }}}}\, + \,3{n_{A{l^{3 + }}}}\, = $ 0,3252 mol (1) ${m_{muoi}} = $ $56.{n_{F{e^{3 + }}}}\, + \,56.{n_{F{e^{2 + }}}}\, + 18.{n_{NH_4^ + }} + \, + \,27{n_{A{l^{3 + }}}}\, + \,35,5.{n_{C{l^ - }}}$ → $56{n_{F{e^{3 + }}}}\, + \,27.{n_{A{l^{3 + }}}} = \,4,3304(2)$ Giải (1), (2) → $\left\{ \begin{array}{l} {n_{F{e^{3 + }}}} = \,0,0484mol\\ {n_{A{l^{3 + }}}} = \,0,06\,mol \end{array} \right.\, \to \,\% {m_{FeC{l_3}}} = \,\frac{{0,0484.162,5}}{{24,2348}}.100 = 32,453\% $