Hỗn hợp X gồm Fe2O3, FeO và...

A

- Hướng dẫn giải

\(\underbrace {\overbrace {{\rm{F}}{{\rm{e}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}}^{{\rm{a}}\,{\rm{mol}}}{\rm{,}}\mathop {{\rm{FeO}}}\limits^{{\rm{b}}\,{\rm{mol}}} {\rm{,Cu}}}_{{\rm{m}}\,\left( {{\rm{gam}}} \right)}\left\langle {\underbrace {{\rm{F}}{{\rm{e}}^{{\rm{2 + }}}}{\rm{,C}}{{\rm{u}}^{{\rm{2 + }}}}{\rm{C}}{{\rm{l}}^{\rm{ - }}}{\rm{,}}\mathop {{{\rm{H}}^{\rm{ + }}}}\limits^{^{{\rm{C}}{{\rm{u}}_{{\rm{du}}\,}}{\rm{:}}\,\,\,{\rm{0,27}}\,\,{\rm{m}}\left( {\rm{g}} \right)}} }_{{\rm{dung}}\,\,{\rm{dich}}\,\,{\rm{Y}}}} \right.\underbrace {{\rm{Ag,AgCl}}}_{{\rm{165,1}}\,\left( {{\rm{gam}}} \right)}{\rm{ + }}\mathop {{\rm{NO}}}\limits_{{\rm{c}}\,\,{\rm{mol}}} \)

Xét hỗn hợp kết tủa ta có: n_AgCl = n_HCl = 1 mol => nAgCl 

\(\frac{{{\rm{m - 143,5nAgCl}}}}{{{\rm{108}}}}{\rm{ = }}\frac{{{\rm{165,1 - 143,4}}{\rm{.1}}}}{{{\rm{108}}}}{\rm{ = 0,2}}\,\,{\rm{mol}}\)

Khi cho X tác dụng với HCl và dung dịch Y tác dụng vói AgNO₃ thì ta có hệ sau:

\(\begin{array}{l}\left\{ \begin{array}{l}{\rm{160}}{{\rm{n}}_{{\rm{F}}{{\rm{e}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}}}{\rm{ + 72}}{{\rm{n}}_{{\rm{FeO}}}}{\rm{ + 64}}{{\rm{n}}_{{\rm{Cupu}}}}{\rm{ = m - }}{{\rm{m}}_{{\rm{Cudu}}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{{{\rm{m}}_{\rm{o}}}}}{{{{\rm{m}}_{\rm{X}}}}}{\rm{ = 0,16}}\\{{\rm{n}}_{{\rm{FeO}}}}{\rm{ + 2}}{{\rm{n}}_{{\rm{Cupu}}}}{\rm{ = 3}}{{\rm{n}}_{{\rm{NO}}}}{\rm{ + nAg}}\\{{\rm{N}}_{{\rm{HCl}}}}{\rm{ = 6}}{{\rm{n}}_{{\rm{F}}{{\rm{e}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}}}{\rm{ + 2}}{{\rm{n}}_{{\rm{FeO}}}}{\rm{ + 4}}{{\rm{n}}_{{\rm{NO}}}}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}{\rm{160a + 72b + 64a = 0,73m}}\left( {\rm{1}} \right)\\\frac{{{\rm{16}}{\rm{.3a + 16b}}}}{{\rm{m}}}{\rm{ = 0,16}}\left( {\rm{2}} \right)\\{\rm{b + 2a = 3c + 0,2}}\\{\rm{6a + 2b + 4c = 1}}\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}5a = b\\b + 2a = 3c + 0,2\\6a + 2b + 4c = 1\end{array} \right. \Rightarrow \left\{ \begin{array}{l}a = 0,05\\b = 0,25\\c = 0,05\end{array} \right.m = 40\end{array}\)