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

* Hướng dẫn giải

Đáp án C

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

Ta có các phản ứng sau:

\(\begin{array}{l}
12{H^ + } + 2NO_3^ – + 10e \to {N_2} + 6{H_2}O\\
0,24\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,02\,\,\,\,0,12
\end{array}\)

\(\begin{array}{l}
10{H^ + } + NO_3^ – + 8e \to NH_4^ + + 3{H_2}O\\
10a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,3a\,\,\,\,\,\,mol
\end{array}\)

\(\begin{array}{l}
2{H^ + } + 2e \to {H_2} \uparrow \\
0,02\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,01\,\,\,\,mol
\end{array}\)

 \( \to {n_{{H_2}O}} = 0,12 + 3a\) 

Bảo toàn khối lượng:

\(\begin{array}{l}
11,2468 + 36,5.(0,26 + 10a) = 24,2348 + 0,02.28 + 0,01.2 + 18.(0,12 + 3a)\\
\to a = 0,02 \to {n_{C{l^ – }}} = 0,26 + 10a = 0,46 = {n_{AgCl}}\\
{n_{Ag}} = \frac{{72,2092 – 0,46.143,5}}{{108}} = 0,0574\,\,(mol)\\
F{e^{2 + }} + A{g^ + } \to Ag + F{e^{3 + }}\\
\to {n_{F{e^{2 + }}}} = {n_{Ag}} = 0,0574\,(mol)
\end{array}\)

Bảo toàn điện tích dung dịch Y:  \(2{n_{F{e^{2 + }}}} + 3{n_{F{e^{3 + }}}} + {n_{NH_4^ + }} + 3{n_{A{l^{3 + }}}} = {n_{C{l^ – }}}\) 

\(\begin{array}{l}
\to 3{n_{F{e^{3 + }}}} + 3{n_{A{l^{3 + }}}} = 0,3252\,\,mol\left( 1 \right)\\
{m_{muoi}} = 56.{n_{F{e^{3 + }}}} + 56.{n_{F{e^{2 + }}}} + 18.{n_{NH_4^ + }} + 27.{n_{A{l^{3 + }}}} + 35,5.{n_{C{l^ – }}}\\
\to 56.{n_{F{e^{3 + }}}} + 27.{n_{A{l^{3 + }}}} = 4,3304\left( 2 \right)
\end{array}\)

Giải (1), (2) \( \to \left\{ \begin{array}{l}
{n_{F{e^{3 + }}}} = 0,0484\\
{n_{A{l^{3 + }}}} = 0,06
\end{array} \right.(mol) \to \% {m_{FeC{l_3}}} = \frac{{0,0484.162,5}}{{24,2348}}.100\% = 32,453\% \)