Hòa tan hoàn toàn 10,8 gam hỗn hợp X gồm Mg, Fe

A

- Hướng dẫn giải

\(X\left\{ \begin{array}{l}Mg:{\rm{ }}a\\F{e_3}{O_4}:{\rm{ }}b\\\underbrace {Cu{{\left( {N{O_3}} \right)}_2})}_{10,8gam}\end{array} \right. \to \left\langle \begin{array}{l}\left\{ \begin{array}{l}NO:{\rm{ }}0,03\\{H_2}:{\rm{ }}0,01\end{array} \right.\\Y\left\{ \begin{array}{l}N{H_4}^ + :{\rm{ }}0,01\\M{g^{2 + }}\\C{u^{2 + }}\\F{e^{2 + }}:{\rm{ }}x\\F{e^{3 + }}:{\rm{ }}y\\SO_4^{2 - }:{\rm{ }}c\end{array} \right. \to \left\langle \begin{array}{l}N{H_3}:{\rm{ }}0,01\\\left\{ \begin{array}{l}Mg,{\rm{ }}Fe,{\rm{ }}Cu\\\underbrace {OH:2c - 0,01}_{13,67{\rm{ }}gam}\end{array} \right.\end{array} \right.\end{array} \right.\)

\(BTNT \to {n_{Cu{{\left( {N{O_3}} \right)}_2}}} = 0,02\)

\(\left\{ \begin{array}{l}{m_x} = 24a + 232b + 188.0,02 = 10,8\\{n_{{H^ + }}} = 2c = 8b + 4.0,03 + 2.0,01 + 10.0,01\\{m_ \downarrow } = 24a + 56.3b + 64.0,02 + 17.(2c - 0,01) = 13,67\end{array} \right. \to \left\{ \begin{array}{l}a = 0,1\\b = 0,02\\c = 0,2\end{array} \right.\)

\(\begin{array}{l}\left\{ \begin{array}{l} \to x + y = 0,06\\ \to 2x + 3y = 0,15\end{array} \right. \to \left\{ \begin{array}{l}x = 0,03\\y = 0,03\end{array} \right.\\ \to C{\% _{F{e_2}{{\left( {S{O_4}} \right)}_3}}} = \frac{{0,015.400}}{{10,8 + 100 - 30.0,03 - 2.0,01}}.100\% = 5,46\% \end{array}\)