Hoà tan hoàn toàn 11,2648 gam hỗn hợp X gồm Fe, Fe(NO3)2 và Al vào dung dịch HCl vừa đủ. Sau phản ứng hoàn toàn, thu được dung dịch Y có chứa 24,2348 gam muối

$\underset{11,2648\left(gam\right)}{\underbrace{X\left\{\begin{array}{l}Fe\\ Fe{\left(N{O}_3\right)}_2\\ Al\end{array}\right.}}\stackrel{+HCl}{\to }\left\{\begin{array}{l}Y\left\{\begin{array}{l}F{e}^{2+}\\ F{e}^{3+}\\ N{H}_4^{+}:a\left(mol\right)\stackrel{+AgN{O}_3}{\to }\downarrow \underset{(72,2092gam)}{\underbrace{\left\{\begin{array}{l}AgCl\\ Ag\end{array}\right.}}\\ A{l}^{3+}\\ C{l}^{-}\end{array}\right.\\ Z\left\{\begin{array}{l}{N}_2:0,02\\ H_2:0,01\end{array}\right.\left(mol\right)+H_{2}O\end{array}\right.$Ta có các phản ứng sau: $\begin{array}{l}12H^{+}+2N{O}_3^{-}+10e\to N_2+6H_{2}O\\ 0,240,020,12\end{array}$

$\begin{array}{l}10H^{+}+N{O}_3^{-}+8e\to N{H}_4^{+}+3H_{2}O\\ 10a a3amol\end{array}$

$\begin{array}{l}2H^{+}+2e\to H_2\uparrow \\ 0,020,01 mol\end{array}$

$\to n_{HCl}=0,26+10a=n_{C{l}^{-}}=n_{AgCl}$

$\to n_{H_{2}O}=0,12+3a$

Bảo toàn khối lượng: $11,2468+36,5.(0,26+10a)=24,2348+0,02.28+0,01.2+18.(0,12+3a)$

$\to a=0,02\to n_{C{l}^{-}}=0,26+10a=0,46=n_{AgCl}$

$n_{Ag}=\frac{72,2092-0,46.143,5}{108}=0,0574\left(mol\right)$

$\begin{array}{l}F{e}^{2+}+A{g}^{+}\to Ag+F{e}^{3+}\\ \to n_{F{e}^{2+}}=n_{Ag}=0,0574\left(mol\right)\end{array}$

Bảo toàn điện tích dung dịch Y: $2n_{F{e}^{2+}}+3n_{F{e}^{3+}}+n_{N{H}_4^{+}}+3n_{A{l}^{3+}}=n_{C{l}^{-}}$

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

$m_{muoi}=56.n_{F{e}^{3+}}+56.n_{F{e}^{2+}}+18.n_{N{H}_4^{+}}+27.n_{A{l}^{3+}}+35,5.n_{C{l}^{-}}$

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

Giải (1), (2) $\to \left\{\begin{array}{l}{n}_{F{e}^{3+}}=0,0484\\ n_{A{l}^{3+}}=0,06\end{array}\right.\left(mol\right)\to \%m_{FeC{l}_3}=\frac{0,0484.162,5}{24,2348}.100\%=32,453\%$