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

Đáp án C

\(\underbrace {X\left\{ \begin{array}{l} Fe\\ Fe{(N{O_3})_2}\\ Al \end{array} \right.}_{11,2648(gam)}\left\{ \begin{array}{l} Y\left\{ \begin{array}{l} F{e^{2 + }}\\ F{e^{3 + }}\\ NH_4^ + :a{\mkern 1mu} {\mkern 1mu} (mol) \downarrow \underbrace {\left\{ \begin{array}{l} AgCl\\ Ag \end{array} \right.}_{(72,2092{\kern 1pt} {\kern 1pt} gam)}\\ A{l^{3 + }}\\ C{l^ - } \end{array} \right.\\ Z\left\{ \begin{array}{l} {N_2}:0,02\\ {H_2}:0,01 \end{array} \right.(mol) + {H_2}O \end{array} \right.\)

Ta có các phản ứng sau:

\(\begin{align} & 12{{H}^{+}}+2NO_{3}^{-}+10e\to {{N}_{2}}+6{{H}_{2}}O \\ & 0,24\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,02\,\,\,\,0,12 \\ \end{align}\)

\(\begin{align} & 10{{H}^{+}}+NO_{3}^{-}+8e\to NH_{4}^{+}+3{{H}_{2}}O \\ & 10a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,3a\,\,\,\,\,\,mol \\ \end{align}\)

\(\begin{align} & 2{{H}^{+}}+2e\to {{H}_{2}}\uparrow \\ & 0,02\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,01\,\,\,\,mol \\ \end{align}\)

\(\to {{n}_{HCl}}=0,26+10a={{n}_{C{{l}^{-}}}}={{n}_{AgCl}}\)

\(\to {{n}_{{{H}_{2}}O}}=0,12+3a\)

Bảo toàn khối lượng:

\(11,2468+36,5.(0,26+10a)=24,2348+0,02.28+0,01.2+18.(0,12+3a)\)

\(\to a=0,02\to {{n}_{C{{l}^{-}}}}=0,26+10a=0,46={{n}_{AgCl}}\)

\({{n}_{Ag}}=\frac{72,2092-0,46.143,5}{108}=0,0574\,\,(mol)\)

\(\begin{align} & F{{e}^{2+}}+A{{g}^{+}}\to Ag+F{{e}^{3+}} \\ & \to {{n}_{F{{e}^{2+}}}}={{n}_{Ag}}=0,0574\,(mol) \\ \end{align}\)

Bảo toàn điện tích dung dịch Y: \(2{{n}_{F{{e}^{2+}}}}+3{{n}_{F{{e}^{3+}}}}+{{n}_{NH_{4}^{+}}}+3{{n}_{A{{l}^{3+}}}}={{n}_{C{{l}^{-}}}}\)

\(\to 3{{n}_{F{{e}^{3+}}}}+3{{n}_{A{{l}^{3+}}}}=0,3252\,\,mol\left( 1 \right)\)

\({{m}_{muoi}}=56.{{n}_{F{{e}^{3+}}}}+56.{{n}_{F{{e}^{2+}}}}+18.{{n}_{NH_{4}^{+}}}+27.{{n}_{A{{l}^{3+}}}}+35,5.{{n}_{C{{l}^{-}}}}\)

\(\to 56.{{n}_{F{{e}^{3+}}}}+27.{{n}_{A{{l}^{3+}}}}=4,3304\left( 2 \right)\)

Giải (1), (2) \(\to \left\{ \begin{align} & {{n}_{F{{e}^{3+}}}}=0,0484 \\ & {{n}_{A{{l}^{3+}}}}=0,06 \\ \end{align} \right.(mol)\to %{{m}_{FeC{{l}_{3}}}}=\frac{0,0484.162,5}{24,2348}.100%=32,453%\)