Hòa tan hoàn toàn 10,8 gam hỗn hợp X gồm Mg, Fe3O4, Cu(NO3)2 trong 100 gam dung dịch H2SO4 (loãng), thu được dung dịch Y chỉ ch�

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

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

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

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