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

$\underbrace{X\left\{ \begin{aligned}
& Fe \\
& Fe{{(N{{O}_{3}})}_{2}} \\
& Al \\
\end{aligned} \right.}_{11,2648(gam)}\xrightarrow{+HCl}\left\{ \begin{aligned}
& Y\left\{ \begin{aligned}
& F{{e}^{2+}} \\
& F{{e}^{3+}} \\
& NH_{4}^{+}:a (mol)\xrightarrow{+AgN{{O}_{3}}}\downarrow \underbrace{\left\{ \begin{aligned}
& AgCl \\
& Ag \\
\end{aligned} \right.}_{(72,2092 gam)} \\
& A{{l}^{3+}} \\
& C{{l}^{-}} \\
\end{aligned} \right. \\
& Z\left\{ \begin{aligned}
& {{N}_{2}}:0,02 \\
& {{H}_{2}}:0,01 \\
\end{aligned} \right.(mol)+{{H}_{2}}O \\
\end{aligned} \right.$
Ta có các phản ứng sau:
$\begin{aligned}
& 12{{H}^{+}}+2NO_{3}^{-}+10e\to {{N}_{2}}+6{{H}_{2}}O \\
& 0,24 0,02 0,12 \\
\end{aligned}$
x$\begin{aligned}
& 10{{H}^{+}}+NO_{3}^{-}+8e\to NH_{4}^{+}+3{{H}_{2}}O \\
& 10a a 3a mol \\
\end{aligned}$
$\begin{aligned}
& 2{{H}^{+}}+2e\to {{H}_{2}}\uparrow \\
& 0,02 0,01 mol \\
\end{aligned}$
$\to {{n}_{HCl}}=0,26+10a={{n}_{C{{l}^{-}}}}={{n}_{AgCl}}$
$\to {{n}_{{{H}_{2}}O}}=0,12+3a$
Bảo toàn khối lượng:
$11,2468+36,5.(0,26+10a)=24,2348+0,02.28+0,01.2+18.(0,12+3a)$
$\to a=0,02\to {{n}_{C{{l}^{-}}}}=0,26+10a=0,46={{n}_{AgCl}}$
${{n}_{Ag}}=\dfrac{72,2092-0,46.143,5}{108}=0,0574 (mol)$
$\begin{aligned}
& F{{e}^{2+}}+A{{g}^{+}}\to Ag+F{{e}^{3+}} \\
& \to {{n}_{F{{e}^{2+}}}}={{n}_{Ag}}=0,0574 (mol) \\
\end{aligned}$
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) $\to \left\{ \begin{aligned}
& {{n}_{F{{e}^{3+}}}}=0,0484 \\
& {{n}_{A{{l}^{3+}}}}=0,06 \\
\end{aligned} \right.(mol)\to \%{{m}_{FeC{{l}_{3}}}}=\dfrac{0,0484.162,5}{24,2348}.100\%=32,453\%$