Phương pháp giải:

+ Từ khối lượng kết tủa (dd Y + AgNO₃) => Fe²⁺, H⁺ trong Y

BTĐT => Fe³⁺ trong Y

+ BTNT H => n_H2O

+ BTKL => m_(NO+N2O) => n_NO, n_N2O

+ BTNT N => n_Fe(NO3)2

+ Đặt ẩn mol Fe và Fe₃O₄ và lập hệ phương trình:

\( \to \left\{ \begin{gathered}
\xrightarrow{{BT:Fe}}{n_{Fe}} + 3{n_{F{e_3}{O_4}}} + {n_{Fe{{(N{O_3})}_2}}} = {n_{F{e^{2 + }}}} + {n_{F{e^{3 + }}}} \hfill \\
{m_{Fe}} + {m_{F{e_3}{O_4}}} + {m_{Fe{{(N{O_3})}_2}}} = {m_X} \hfill \\
\end{gathered} \right. \to \left\{ \begin{gathered}
x = ? \hfill \\
y = ? \hfill \\
\end{gathered} \right.\)

Lời giải chi tiết:

Do dd Y tác dụng với AgNO₃ dư thu được NO nên dd Y chứa Fe²⁺, Fe³⁺, Cl^-, H⁺ dư.

\(\begin{gathered}
18,72g\left\{ \begin{gathered}
Fe:x \hfill \\
F{e_3}{O_4}:y \hfill \\
Fe{(N{O_3})_2} \hfill \\
\end{gathered} \right\}\xrightarrow{\begin{subarray}{l}
+ HCl:0,75 \\
+ HNO3:0,06
\end{subarray} }\left| \begin{gathered}
ddY\left\{ \begin{gathered}
F{e^{2 + }}:0,09 \hfill \\
\xrightarrow{{BTDT}}F{e^{3 + }}:0,16 \hfill \\
C{l^ - }:0,75 \hfill \\
{H^ + }du:0,09 \hfill \\
\end{gathered} \right\}\xrightarrow{{ + AgNO3du}}\left| \begin{gathered}
\left\{ \begin{gathered}
AgCl:0,75 \hfill \\
\xrightarrow{{m \downarrow }}Ag:0,0225 \hfill \\
\end{gathered} \right\} \hfill \\
\hfill \\
NO:0,0225 \hfill \\
\left( \begin{gathered}
F{e^{2 + }} + \,\,\,\,\,\,\,\,4{H^ + } + N{O_3}^ - \to 3F{e^{3 + }} + NO + 2{H_2}O \hfill \\
0,0675 \leftarrow 0,09\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \leftarrow 0,0225 \hfill \\
F{e^{2 + }} + A{g^ + } \to F{e^{3 + }} + Ag \hfill \\
0,0225\,\,\,\,\,\,\,\,\,\, \leftarrow \,\,\,\,\,\,\,\,\,\,\,\,\,0,0225 \hfill \\
\end{gathered} \right) \hfill \\
\end{gathered} \right. \hfill \\
\left\{ \begin{gathered}
NO:a \hfill \\
{N_2}O:b \hfill \\
\end{gathered} \right. \hfill \\
\hfill \\
{H_2}O \hfill \\
\end{gathered} \right. \hfill \\
\xrightarrow{{BT:H}}m{H_2}O = \frac{{{n_{HCl}} + {n_{HN{O_3}}} - {n_{{H^ + }du}}}}{2} = 0,36mol \hfill \\
\xrightarrow{{BTKL}}{m_{(NO + {N_2}O)}} = {m_X} + {m_{HCl}} + {m_{HN{O_3}}} - {m_Y} - {m_{{H_2}O}} = 2,68g \hfill \\
\to \left\{ \begin{gathered}
a + b = 0,08 \hfill \\
30a + 44b = 2,68 \hfill \\
\end{gathered} \right. \to \left\{ \begin{gathered}
a = 0,06 \hfill \\
b = 0,02 \hfill \\
\end{gathered} \right. \hfill \\
\xrightarrow{{BT:N}}{n_{Fe{{(N{O_3})}_2}}} = \frac{{{n_{NO}} + 2{n_{{N_2}O}} - {n_{HN{O_3}}}}}{2} = 0,02mol \hfill \\
\to \left\{ \begin{gathered}
\xrightarrow{{BT:Fe}}x + 3y + 0,02 = 0,09 + 0,16 \hfill \\
56x + 232y + 0,02.180 = 18,72( = mX) \hfill \\
\end{gathered} \right. \to \left\{ \begin{gathered}
x = 0,125 \hfill \\
y = 0,035 \hfill \\
\end{gathered} \right. \hfill \\
\to {m_{F{e_3}{O_4}}} = 0,035.232 = 8,12g \hfill \\
\end{gathered} \)

Đáp án D