Hoà tan hết 17,91 gam hỗn hợp X gồm Fe, FeCO3, Zn và ZnCO3 trong...

$\begin{aligned}
& \xrightarrow{Quy \tilde{n}oi X}\underbrace{\left\{ \begin{aligned}
& Mg \\
& Zn \\
& Fe \\
& C{{O}_{3}}:x \\
\end{aligned} \right.}_{17,91 gam}+\underbrace{\left\{ \begin{aligned}
& NaHS{{O}_{4}}:0,62 \\
& HN{{O}_{3}}:0,08 \\
\end{aligned} \right.}_{79,44 gam}\xrightarrow{{}}\left| \begin{aligned}
& m gam\left\{ \begin{aligned}
& \left. \left\{ \begin{aligned}
& M{{g}^{2+}};Z{{n}^{2+}}; \\
& F{{e}^{2+}};F{{e}^{3+}} \\
\end{aligned} \right. \right\}:(17,91-60x) gam \\
& NH_{4}^{+}:y \\
& \left. \left\{ \begin{aligned}
& N{{a}^{+}}:0,62 \\
& SO_{4}^{2-}:0,62 \\
\end{aligned} \right. \right\}73,78 gam \\
\end{aligned} \right. \\
& \\
& 0,19 mol \uparrow Y\left\{ \begin{aligned}
& C{{O}_{2}}:x \\
& NO:(0,08-y) (BT.N) \\
& {{H}_{2}}:0,03 \\
\end{aligned} \right. \\
& {{H}_{2}}O:(0,32-2y) (BT.H) \\
\end{aligned} \right. \\
& \\
\end{aligned}$
$\to \left\{ \begin{aligned}
& {{n}_{Y}}=x+(0,08-y)+0,03=0,19 \\
& \xrightarrow{BTKL}60x+79,44=(18y+73,78)+[44x+30(0,08-y)+2.0,03]+18(0,32-2y) \\
\end{aligned} \right.$
$\to \left\{ \begin{aligned}
& x=0,1 \\
& y=0,02 \\
\end{aligned} \right.\to m=(17,91-60.0,1)+18.0,02+73,78=86,05 gam$