Hòa tan hoàn toàn m gam hỗn hợp X chứa Cu, Mg...

$m(g)X\left\{ \begin{aligned}
& Cu \\
& Mg \\
& F{{e}_{3}}{{O}_{4}} \\
& Fe{{\left( N{{O}_{3}} \right)}_{2}} \\
\end{aligned} \right.\xrightarrow{HCl:0,61(mol)}\left\langle \begin{aligned}
& \left( m+16,195 \right)(g)Y\left\{ \begin{aligned}
& F{{e}^{2+}} \\
& M{{g}^{2+}} \\
& C{{u}^{2+}} \\
& NH_{4}^{+} \\
& C{{l}^{-}} \\
\end{aligned} \right.\xrightarrow{NaOH}24,44(g)\downarrow \\
& 0,085(mol)Z\left\{ \begin{aligned}
& NO \\
& {{H}_{2}} \\
\end{aligned} \right.m=1,57(g)+{{H}_{2}}O \\
\end{aligned} \right.$
$BTKL:m+0,6.36,5=\left( m+16,195 \right)+1,57+{{m}_{{{H}_{2}}O}}\to {{m}_{{{H}_{2}}O}}=4,5(g)\to {{n}_{{{H}_{2}}O}}=0,25(mol)$
$Z\left\{ \begin{aligned}
& NO:a \\
& {{H}_{2}}:b \\
\end{aligned} \right.\to \left\{ \begin{aligned}
& a+b=0,085 \\
& 30a+2b=1,57 \\
\end{aligned} \right.\to \left\{ \begin{aligned}
& a=0,05 \\
& b=0,035 \\
\end{aligned} \right.$
BTNT (Cl): ${{n}_{C{{l}^{-}}(Y)}}=0,61(mol)$
BTNT (H): ${{n}_{NH_{4}^{+}(Y)}}=\dfrac{{{n}_{HCl}}-2{{n}_{{{H}_{2}}O}}-2{{n}_{{{H}_{2}}}}}{4}=0,01(mol)$
BTNT (N): ${{n}_{Fe{{\left( N{{O}_{3}} \right)}_{2}}}}=\dfrac{{{n}_{NO}}+{{n}_{NH_{4}^{+}(Y)}}}{2}=0,03(mol)$
BTNT (O): ${{n}_{F{{e}_{3}}{{O}_{4}}}}=\dfrac{{{n}_{NO}}+{{n}_{{{H}_{2}}O}}-6{{n}_{Fe{{\left( N{{O}_{3}} \right)}_{2}}}}}{4}=0,03(mol)$
$\to \sum{{{n}_{Fe}}}=0,03+0,03.3=0,12(mol)$
$\left\{ \begin{aligned}
& M{{g}^{2+}}:x \\
& C{{u}^{2+}}:y \\
\end{aligned} \right.\to \left\{ \begin{aligned}
& 2a+2y+2{{n}_{F{{e}^{2+}}}}+{{n}_{NH_{4}^{+}(Y)}}=0,61 \\
& {{m}_{kt}}={{m}_{Mg{{(OH)}_{2}}}}+{{m}_{Cu{{(OH)}_{2}}}}+{{m}_{Fe{{(OH)}_{2}}}}=24,4 \\
\end{aligned} \right.\to \left\{ \begin{aligned}
& x=0,1 \\
& y=0,08 \\
\end{aligned} \right.$
$\to \%=\dfrac{0,08.64.100\%}{0,08.64+0,1.24+0,03.232+0,03.180}=25,75\%$