Hỗn hợp X gồm Fe2O3, FeO và Cu (trong đó sắt chiếm 52,5% về khối lượng). Cho m

$m(g)X\left\{ \begin{array}{l}
F{e_2}{O_3}\\
FeO\\
Cu
\end{array} \right.\, + 0,84\,mol\,HCl\, \to \,\left| {{\rm{dd}}\,Y\,\left\{ \begin{array}{l}
F{e^{2 + }}:\,x\,mol\\
C{u^{2 + }}:\,y\,mol\\
C{l^ - }:\,0,84\,mol\\
{H^ + }_{du}:\,0,84 - 2x - 2y\,mol\,\\
{H_2}O:\,z\,mol\\
0,2m\,C{u_{du}}
\end{array} \right.} \right.\,$ $ + \,AgN{O_{3\,du}}\, \to \,\left| \begin{array}{l}
F{e^{3 + }},\,C{u^{2 + }}\\
NO\\
141,6(g)\,\left\{ \begin{array}{l}
AgCl:\,0,84\,mol\\
Ag
\end{array} \right.
\end{array} \right.$
${n_{Ag}}\,\frac{{141,6 - 0,84.143,5}}{{108}} = 0,195mol$
Quy đổi : 0,8m (g) hỗn hợp X $\left\{ \begin{array}{l}
Fe:\,x\,mol\\
Cu:y\,mol\\
O:\,z\,mol
\end{array} \right.$ →${m_{hh\,X}} = \,\frac{{56x + 64y + 16z}}{{0,8}}$
Ta có : $\% {m_{Fe(X)}} = \,\frac{{56x}}{{\frac{{56x + 64y + 16z}}{{0,8}}}} = \,\frac{{52,5}}{{100}}(1)$
Bảo toàn nguyên tố H → ${n_{HCl}} = \,{n_{{H^ + }({\rm{dd}}\,Y)}}\, + \,{n_{{H_2}O}} \to \,0,84 = 0,84 - 2x - 2y + 2z(2)$
Các quá trình trao đổi electron
$\underbrace {Fe\,}_x \to \,F{e^{3 + }}\, + \underbrace {\,3e}_x\,$
$\underbrace {Cu}_y\, \to \,C{u^{2 + }}\, + \,\underbrace {2e}_{2y}$
$\underbrace {O\,}_z + \underbrace {\,2e}_{2z} \to {O^{ - 2}}$
$\underbrace {4{H^ + }}_{0,84 - 2x - 2y}\,\, + NO_3^ - + 3e \to \underbrace {NO}_{\frac{3}{4}(0,84 - 2x - 2y)}\, + \,2{H_2}O$
$Ag\, + 1e \to \underbrace {Ag}_{0,195}$
Bảo toàn e d 3x+ 2y = 2z + 0,75. (0,84-2x-2y) + 0,196 (3)
Từ (1), (20, (3) → x = 0,3 ; y = 0,05 ; z = 0,35
→ m = $\frac{{56x + 64y + 16z}}{{0,8}} = 32$