Hòa tan hết 37,28 gam hỗn hợp X gồm Fe3O4, Cu trong 500 ml dung dịch chứa HCl

$37,28(g)\,X\left\{ \begin{array}{l}
F{e_3}{O_4}:x\,mol\\
Cu:\,y\,mol
\end{array} \right.\,\, + \,\left\{ \begin{array}{l}
HCl:\,1,2\,mol\\
HN{O_3}:\,0,1\,mol
\end{array} \right.\,\, \to \,\left| \begin{array}{l}
{\rm{dd}}\,Y\,\left\{ \begin{array}{l}
F{e^{2 + }}\\
F{e^{3 + }}\\
C{u^{2 + }}\\
C{l^ - }\\
NO_3^ -
\end{array} \right.\,\,\left[ \begin{array}{l}
+ \,AgN{O_3}\, \to \,\, \downarrow \left\{ \begin{array}{l}
AgCl\\
Ag
\end{array} \right.\\
+ NaOH\, \to \downarrow \, \to \,\left\{ \begin{array}{l}
F{e_2}{O_3}\\
CuO
\end{array} \right.
\end{array} \right.\\
NO\, + \,{H_2}O
\end{array} \right.$
Ta có: $\left\{ \begin{array}{l}
232x + 64y = 37,28\\
160.\frac{{3x}}{2} + 80y = 41,6
\end{array} \right. \to \left\{ \begin{array}{l}
x = 0,1 = \,{n_{F{e_3}{O_4}}}\\
y = 0,22 = {n_{Cu}}
\end{array} \right.$
Các quá trình xảy ra
$\underbrace {Fe_3^{\frac{{ + 8}}{3}}}_{0,1}\,\, \to \,3F{e^{ + 3}}\,\, + \underbrace {1e}_{0,1}$
$\,\underbrace {C{u^0}}_{0,22}\,\, \to \,C{u^{2 + }}\, + \,\,\underbrace {2e}_{0,44}$
$\,\underbrace {{O^{2 - }}}_{0,4}\, + \,\,\underbrace {2{H^ + }}_{0,8}\,\, \to {H_2}O$
$\,\underbrace {4{H^ + }}_{1,3 - 0,8} + \,NO_3^ - \,\, + \,\,\underbrace {3e}_{0,375} \to NO\, + \,2{H_2}O$
$A{g^ + }\,\, + \,\,\underbrace {1e}_a\, \to \,\,\underbrace {Ag}_a$
Bảo toàn e → 0,1 + 0,44 = 0,375 +a
→ a = 0,165 = ${n_{Ag}}$
→ ${m_ \downarrow }$ = ${m_{AgCl}}\, + \,{m_{Ag}}$ = 1,2 . 143,5 + 0,165 . 108 = 190, 02 (g)