Phương pháp giải:

BTNT, BTKL.

Lời giải chi tiết:

\(\begin{gathered}
\underbrace {\left\{ \begin{gathered}
FeO:x \hfill \\
F{e_3}{O_4}:y \hfill \\
Cu:z \hfill \\
\end{gathered} \right.}_{27,36g} + \underbrace {NaN{O_3}}_{0,04(BTNT:N)} + HCl \to \underbrace {Muoi\,clorua}_{58,16g} + \underbrace {NO}_{0,04mol} + {H_2}O \hfill \\
\to x = \frac{1}{4}(x + y + z)\,(1) \hfill \\
\to 72x + 232y + 64z = 27,36\,(2) \hfill \\
\xrightarrow{{BTNT\,O}}x + 4y + 0,04.3 = 0,04 + {n_{{H_2}O}} \to {n_{{H_2}O}} = x + 4y + 0,08 \hfill \\
\xrightarrow{{BTNT\,H}}{n_{HCl}} = 2{n_{{H_2}O}} = 2x + 8y + 0,16 \hfill \\
\xrightarrow{{BTKL}}27,36 + 85.0,04 + 36,5(2x + 8y + 0,16) = 58,16 + 0,04.30 + 18(x + 4y + 0,08)\,(3) \hfill \\
\xrightarrow{{(1)(2)(3)}}\left\{ \begin{gathered}
x = 0,04 \hfill \\
y = 0,1 \hfill \\
z = 0,02 \hfill \\
\end{gathered} \right. \to \left\{ \begin{gathered}
Fe:0,34 \hfill \\
Cu:0,02 \hfill \\
O:0,44 \hfill \\
\end{gathered} \right. \hfill \\
{n_{F{e^{2 + }}}} = a;\,\,{n_{F{e^{3 + }}}} = b \hfill \\
\to \left\{ \begin{gathered}
a + b = {n_{Fe}} = 0,34 \hfill \\
2a + 3b + 0,02.2 = 0,44.2 + 0,04.3(BTe) \hfill \\
\end{gathered} \right. \to \left\{ \begin{gathered}
a = 0,06 \hfill \\
b = 0,28 \hfill \\
\end{gathered} \right. \hfill \\
\to \left\{ \begin{gathered}
{n_{AgCl}} = {n_{HCl}} = 2x + 8y + 0,16 = 1,04 \hfill \\
{n_{Ag}} = {n_{F{e^{2 + }}}} = 0,06 \hfill \\
\end{gathered} \right. \to {m_ \downarrow } = {m_{AgCl}} + {m_{Ag}} = 1,04.143,5 + 0,06.108 = 155,72g \hfill \\
\end{gathered} \)

Đáp án D