Đáp án đúng: B
32.
– Quá trình: $\underbrace{\overbrace{{F}{{{e}}_{{2}}}{{{O}}_{{3}}}}^{{a}\,{mol}}{,}\,\overbrace{{FeO}}^{{b}\,{mol}}{,}\,{Cu}}_{{m}\,{(g)}}\xrightarrow{{HCl}}\left\langle \begin{array}{l}{Cu:}\,{0,2}\,{m}\,{(g)}\\\underbrace{{F}{{{e}}^{{2+}}}{,}\,{C}{{{u}}^{{2+}}}{,}\,{C}{{{l}}^{{-}}}{,}\,{{{H}}^{{+}}}}_{{dd Y}}\xrightarrow{{AgN}{{{O}}_{{3}}}}\underbrace{{Ag,}\,{AgCl}}_{{141,6}\,{(g)}\,\downarrow }{+}\underbrace{{NO}}_{{c}\,{mol}}\end{array} \right.$
– Xét hỗn hợp kết tủa ta có :$\displaystyle \xrightarrow{{BT:}\,{Cl}}{{{n}}_{{AgCl}}}{=}{{{n}}_{{HCl}}}{=0,84}\,{mol}\Rightarrow {{{n}}_{{Ag}}}{=}\frac{{m}\downarrow {-143,5}{{{n}}_{{AgCl}}}}{{108}}{=0,195}\,{mol}$
– Khi cho X tác dụng với HCl và dung dịch Y tác dụng với AgNO3 thì ta có hệ sau :
$\Rightarrow \left\{ \begin{array}{l}{160}{{{n}}_{{F}{{{e}}_{{2}}}{{{O}}_{{3}}}}}{+72}{{{n}}_{{FeO}}}{+64}{{{n}}_{{Cu(pu)}}}{=m-}{{{m}}_{{r}}}\\\frac{{{{m}}_{{Fe}}}}{{{{m}}_{{X}}}}{=0,525}\\\xrightarrow{{BT:}\,{e}}{{{n}}_{{FeO}}}{+2}{{{n}}_{{Cu(pu)}}}{=3}{{{n}}_{{NO}}}{+}{{{n}}_{{Ag}}}\\{{{n}}_{{HCl}}}{=6}{{{n}}_{{F}{{{e}}_{{2}}}{{{O}}_{{3}}}}}{+2}{{{n}}_{{FeO}}}{+4}{{{n}}_{{NO}}}\end{array} \right.$$\Rightarrow \left\{ \begin{array}{l}{160a+72b+64a=0,8m}\\\frac{{56}{.2a+56b}}{{m}}{=0,525}\\{b+2a=3c+0,195}\\{6a+2b+4c=0,84}\end{array} \right.\Rightarrow \left\{ \begin{array}{l}{a=}\,{0,05}\\{b=}\,{0,2}\\{c=}\,{0,035}\\{m=32}\end{array} \right.$
