Cho 66,2 gam hỗn hợp X gồm Fe3O4, Fe(NO3)2, Al tan hoàn toàn trong dung dịch

${d_{Z/{H_2}}} = \,\frac{{23}}{9}$ → ${\overline M _{hh\,Z}}\, = \,\,\frac{{46}}{9}$
Z gồm 2 khí, trong đó có 1 khí hóa nâu ngoài không khí, khí đó là NO (M = 30 > $\overline {{M_Z}} $ )
→ Khí còn lại có M < ${\overline M _Z}\,$ . Đó chỉ có thể là ${H_2}$ (M = 2)
Phương pháp đường chéo → $\left\{ \begin{array}{l}
{n_{NO}} = \,0,05\,\,mol\,\\
{n_{{H_2}}} = \,0,4\,\,mol
\end{array} \right.$
+) Khí Z chứa → hết, hết
${H_2}$ Muối + khí Z + ${H_2}O$
BTKL: ${m_{{H_2}O}}\, = \,66,2\,\, + \,\,3,1.136\,\, - \,466,6\,\, + \,\,0,4.2\,\, + \,\,0,05.30\, = \,18,9$ (g)
$→{n_{KOH}}\, = \,0,6\,mol\,\, \to \,\,\sum {{n_{O{H^ - }}}\,\, = \,\,0,9\,\,mol}$
$C{O_2}\,\,\,\, + \,\,2O{H^ - }\,\,\, \to \,\,CO_3^{2 - }\,\,\,\, + \,\,{H_2}O\\$
$a...............2a..............a$
$C{O_2}\,\, + \,\,O{H^ - }\,\,\, \to \,\,HCO_3^ - \\$
$b.............b$
→ $\left\{ \begin{array}{l}
a\,\,\, + \,\,\,b\,\, = \,\,0,5\,\\
a\,\, + \,\,b\,\, = \,\,0,9\,\,
\end{array} \right.\,\,\, \to \left\{ \begin{array}{l}
a\,\, = \,\,0,1\\
b\,\, = \,\,0,9\,\,
\end{array} \right.\, \to \,\,{n_ \downarrow }\,\, = \,\,x\,\, = \,\,0,1\,\,mol$
Mặt khác, có $\sum {{n_{SO_4^{2 - }}} = \,3,1\,\,mol} $ → ${m_{Fe}}\,\, + \,\,{m_{Al}}\,\, = \,\,47,2$
${n_{NO_3^ - }}\,\, = \,\,{n_{NH_4^ + }}\,\, + \,\,{n_{NO}}\,\, = \,\,0,05\,\, + \,\,0,05\,\, = \,\,0,1\,\,mol$
$ \to \,\,{n_{Fe{{(N{O_3})}_2}}}\,\, = \,\,0,05\,\, \to \,\,{m_{F{e_3}{O_4}\,\,}} + \,\,{m_{Al}}\,\, = \,\,66,2\,\, - \,\,0,05.180\,\, = \,\,57,2$
Đặt $\left\{ \begin{array}{l}
{n_{F{e_3}{O_4}}} = \,\,\,x\\ {n_{Al}} = \,y \end{array} \right.\,\,\, \to \left\{ \begin{array}{l}
232x\,\,\, + \,\,27\,y\,\, = \,\,57,2\\
168x\,\, + \,\,27y\,\, = \,\,47,2 - 0,05.56
\end{array} \right.\,\,
\to \,\,\left\{ \begin{array}{l}
x\,\, = \,\,0,02\\ y\,\, = \,\,0,4 \end{array} \right.$
→ ${m_{Fe}} = \,0,2.3\, + \,\,0,05\, = \,0,65\,\,mol\,\, \to \,\,\% Fe\,\, = \,\,54,98\% $