Bài giải

$2yAl\text{ }+3F{{e}_{x}}{{O}_{y}}\text{ }\xrightarrow{{{t}^{0}}}\text{ }3xFe\text{ }+\text{ }yA{{l}_{2}}{{O}_{3}}\text{ }\left( 1 \right)$

Phần 1 :
$Fe\text{ }+4HN{{O}_{3}}\text{ }\xrightarrow{{}}\text{ }Fe{{\left( N{{O}_{3}} \right)}_{3}}\text{ }+\text{ }NO\text{ }+2{{H}_{2}}O\text{ }\left( 2 \right)$

$A{{l}_{2}}{{O}_{3}}\text{ }+\text{ }6HN{{O}_{3}}\text{ }\xrightarrow{{}}\text{ }2Al{{\left( N{{O}_{3}} \right)}_{3}}\text{ }+3{{H}_{2}}O\text{ }\left( 3 \right)$

$Al\text{ }+4HN{{O}_{3}}\text{ }\to Al{{\left( N{{O}_{3}} \right)}_{3}}\text{ }+NO\text{ }+2{{H}_{2}}O\text{ }\left( 4 \right)$

Phần 2 :

$A{{l}_{2}}{{O}_{3}}\text{ }+2NaOH\text{ }2NaAl{{O}_{2}}\text{ }+{{H}_{2}}O\text{ }\left( 5 \right)$

$2Al\text{ }+2{{H}_{2}}O\text{ }+2NaOH\text{ }\xrightarrow{{}}\text{ }2NaAl{{O}_{2}}\text{ }+3{{H}_{2}}\text{ }\left( 6 \right)$

Vì khi cho phần 2 tác dụng với NaOH dư thấy giải phóng ${{H}_{2}}$ $\to Al$ dư sau phản ứng (1)

${{n}_{{{H}_{2}}}}=0,015\left( mol \right)$

${{n}_{Fe\left( P2 \right)}}=0,045\left( mol \right)$

Giả sử phần 1 gấp k lần phần 2

$\Rightarrow \text{ }{{n}_{Fe\left( P1 \right)}}=0,045k\left( mol \right)$

Theo phương trình (5) : ${{n}_{Al\left( P2 \right)}}=\frac{2}{3}{{n}_{H2}}=0,01\left( mol \right)$

$\Rightarrow {{n}_{Al\left( P1 \right)}}=0,01k\left( mol \right)$

${{n}_{NO}}=0,165\left( mol \right)$

Theo phương trình (2,4) : ${{n}_{NO}}=0,045k+0,01k\text{ }\left( mol \right)$

$\Rightarrow 0,055k=0,165\Rightarrow \text{ }k=3$

$\Rightarrow {{n}_{Al\left( P1 \right)}}=0,03\left( mol \right)$

${{n}_{Fe\left( p1 \right)}}=0,135\left( mol \right)$

$\sum\limits_{{}}^{{}}{{}}{{n}_{Fe}}\text{ }=\text{ }0,045\text{ }+\text{ }0,135\text{ }=\text{ }0,18\text{ }\left( mol \right)$

$\sum\limits_{{}}^{{}}{{}}{{n}_{Al}}\text{ }=\text{ }0,03\text{ }+\text{ }0,01\text{ }=\text{ }0,04\text{ }\left( mol \right)$

${{m}_{Al2O3}}=14,49-0,135.56-0,03.27=6,12\text{ }\left( g \right)$

${{n}_{A{{l}_{2}}{{O}_{3}}\left( P1 \right)}}=0,06\left( mol \right)$

$\Rightarrow \text{ }{{n}_{A{{l}_{2}}{{O}_{3}}\left( P2 \right)}}=0,02\left( mol \right)$

$\Rightarrow {{n}_{A{{l}_{2}}O}}_{_{3}}\text{ }=0,08\text{ }\left( mol \right)$

Theo (1): ${{n}_{Fe}}=\frac{3x}{y}{{n}_{A{{l}_{2}}{{O}_{3}}}}$

$\Rightarrow \text{ }0,18=3.\frac{x}{y}.0,08\Rightarrow \text{ }\frac{x}{y}=\frac{3}{4}$

$\Rightarrow $ CTHH : $F{{e}_{3}}{{O}_{4}}$

Theo (1) : ${{n}_{F{{e}_{3}}{{O}_{4}}}}\text{ }=\frac{1}{3}.{{n}_{Fe}}=0,06\left( mol \right)$

$\Rightarrow m=0,04.27+0,06.232=15\left( g \right)$