• ${{n}_{Al}}=0,02$mol, ${{n}_{F{{e}_{3}}{{O}_{4}}}}=0,006$mol


• PTHH: 8Al + $3F{{e}_{3}}{{O}_{4}}$$\xrightarrow{{{t}^{0}}}$ $4A{{l}_{2}}{{O}_{3}}$ + 9Fe


• Xét tỉ lệ: $\frac{{{n}_{Al}}}{8}$< $\frac{{{n}_{F{{e}_{3}}{{O}_{4}}}}}{3}$ $\Rightarrow $ $F{{e}_{3}}{{O}_{4}}$ phản ứng hết, Al còn dư. Do đó tính theo $F{{e}_{3}}{{O}_{4}}$.


• Theo PTHH ta có: ${{n}_{Al(pu)}}=\frac{8}{3}{{n}_{F{{e}_{3}}{{O}_{4}}}}$= $\frac{8}{3}$.0,006 = 0,016 mol

$\Rightarrow $${{n}_{Al(du)}}$= 0,02-0,016 = 0,004 mol

• Trong $\frac{1}{2}$A có ${{n}_{Al(du)}}=0,002$mol


• PTHH: Al + NaOH + ${{H}_{2}}O$  $\xrightarrow{{}}$NaAl${{O}_{2}}$ + $\frac{3}{2}{{H}_{2}}\uparrow $


• Theo PTHH ta có: ${{n}_{NaOH}}={{n}_{Al(du)}}=0,002$ mol


• Suy ra: V = $\frac{n}{{{C}_{M}}}=\frac{0,002}{1}=0,002$lít = 2 ml


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