Phương pháp giải:

1. Bảo toàn e

2. Bảo toàn điện tích

Lời giải chi tiết:

+ Y chứa HNO₃ dư nên Y không chứa Fe²⁺

\(\left\{ \begin{gathered}
{n_{NO}} + {n_{{N_2}O}} = 0,04 \hfill \\
30{n_{NO}} + 44{n_{{N_2}O}} = 0,04.16,75.2 = 1,34 \hfill \\
\end{gathered} \right. \to \left\{ \begin{gathered}
{n_{NO}} = 0,03 \hfill \\
{n_{{N_2}O}} = 0,01 \hfill \\
\end{gathered} \right.\)

+ Sơ đồ phản ứng:

\(\begin{gathered}
X\left\{ \begin{gathered}
Mg \hfill \\
F{e_3}{O_4} \hfill \\
\end{gathered} \right\}\xrightarrow{{ + HNO3:0,6\,mol}}\left| \begin{gathered}
\to \left\{ \begin{gathered}
NO \hfill \\
{N_2}O \hfill \\
\end{gathered} \right\} \hfill \\
\to Y\left\{ \begin{gathered}
M{g^{2 + }},F{e^{3 + }} \hfill \\
N{H_4}^ + ,N{O_3}^ - \hfill \\
{H^ + } \hfill \\
\end{gathered} \right\}\xrightarrow{{ + NaOH(trung\,hoa)}}Z\left\{ \begin{gathered}
M{g^{2 + }},F{e^{3 + }} \hfill \\
N{H_4}^ + ,N{O_3}^ - \hfill \\
N{a^ + } \hfill \\
\end{gathered} \right\} \hfill \\
\end{gathered} \right. \hfill \\
\hfill \\
\end{gathered} \)

\(\left\{ \begin{gathered}
{n_{Mg}} = x \hfill \\
{n_{F{e_3}{O_4}}} = y \hfill \\
\end{gathered} \right. \to \left\{ \begin{gathered}
{n_{M{g^{2 + }}}} = x \hfill \\
{n_{F{e^{3 + }}}} = 3y \hfill \\
\end{gathered} \right.;\left\{ \begin{gathered}
{n_{N{H_4}^ + }} = z \hfill \\
BTN:{n_{N{O_3}^ - (Z)}} = 0,55 - z \hfill \\
\end{gathered} \right.\)

\(\begin{gathered}
\left\{ \begin{gathered}
{m_X} = 24x + 232y = 9,6 \hfill \\
BT\,e:2x + y = 0,03.3 + 0,01.8 + 8z \hfill \\
BTDT\,{\text{dd}}\,Z:2x + 9y + z + 0,04 = 0,55 - z \hfill \\
\end{gathered} \right. \to \left\{ \begin{gathered}
x = 0,11 \hfill \\
y = 0,03 \hfill \\
z = 0,01 \hfill \\
\end{gathered} \right. \hfill \\
\to {m_{muoi\,trong\,Z}} = {m_{F{e^{3 + }}}} + {m_{M{g^{2 + }}}} + {m_{N{H_4}^ + }} + {m_{N{a^ + }}} + {m_{N{O_3}^ - }} = 42,26(g) \hfill \\
\end{gathered} \)

Đáp án A