Phương pháp giải:

Bảo toàn điện tích, bảo toàn nguyên tố, bảo toàn e

Lời giải chi tiết:

\(\begin{gathered}
13,68(g)\left\{ \begin{gathered}
Fe:x \hfill \\
FeC{O_3}:y \hfill \\
\end{gathered} \right\} + \left\{ \begin{gathered}
{H_2}S{O_4}:0,405 \hfill \\
NaN{O_3}:0,45 \hfill \\
\end{gathered} \right\} \to \left\{ \begin{gathered}
dd\,X\xrightarrow{{ + NaOH}}Fe{(OH)_3}:0,17 \hfill \\
Khi\,Y\left\{ \begin{gathered}
C{O_2}:y \hfill \\
N{O_2} \hfill \\
NO:0,12 \hfill \\
\end{gathered} \right\} \hfill \\
\end{gathered} \right. \hfill \\
+ Ta\,co:56x + 116y = 13,68(1) \hfill \\
+ Ta\,co:{n_{NaOH}} = {n_{{H^ + }}} + 3{n_{Fe{{(OH)}_3}}} \to {n_{{H^ + }}} = 0,09(mol) \hfill \\
X\left\{ \begin{gathered}
F{e^{3 + }}:x + y \hfill \\
N{a^ + }:0,45 \hfill \\
{H^ + }:0,09 \hfill \\
S{O_4}^{2 - }:0,405 \hfill \\
\xrightarrow{{BTDT}}N{O_3}^ - :3x + 3y - 0,27 \hfill \\
\end{gathered} \right\} \hfill \\
\xrightarrow{{BTN}}{n_{N{O_2}}} = {n_{NaN{O_3}}} - {n_{N{O_3}^ - (X)}} - {n_{NO}} = 0,45 - (3x + 3y - 0,27) - 0,12 = 0,6 - 3x - 3y \hfill \\
\xrightarrow{{BTe}}3{n_{Fe}} + {n_{FeC{O_3}}} = {n_{N{O_2}}} + 3{n_{NO}} \to 3x + y = 0,6 - 3x - 3y + 3.0,12(2) \hfill \\
\xrightarrow{{(1)(2)}}\left\{ \begin{gathered}
x = 0,12 \hfill \\
y = 0,06 \hfill \\
\end{gathered} \right. \hfill \\
\to Y\left\{ \begin{gathered}
C{O_2}:y = 0,06 \hfill \\
N{O_2}:0,6 - 3x - 3y = 0,06 \hfill \\
NO:0,12 \hfill \\
\end{gathered} \right. \to {M_Y} = \frac{{0,06.44 + 0,06.46 + 0,12.30}}{{0,06 + 0,06 + 0,12}} = 37,5 \hfill \\
\to {d_{Y/He}} = \frac{{37,5}}{4} = 9,375 \hfill \\
\end{gathered} \)

Đáp án D