- Quá trình : $\underbrace{Al}_{0,2\,mol}+\underbrace{\overbrace{Fe{{(N{{O}_{3}})}_{3}}}^{0,15\,mol},\,\overbrace{Cu{{(N{{O}_{3}})}_{2}}}^{0,15\,mol}}_{\text{dung dch X}}\to \underbrace{\overbrace{A{{l}^{3+}}}^{0,2\,mol},\,F{{e}^{2+}},\overbrace{\,N{{O}_{3}}^{-}}^{0,75\,mol}}_{\text{dung dch sau p}}+\underbrace{Fe,\,Cu}_{m\,(g)\,\text{r }\!\!{\scriptscriptstyle 3\!/\!{ }_4}\!\!\text{ n}}$

$\xrightarrow[dd\text{ sau p}]{BTDT}{{n}_{F{{e}^{2+}}}}=\frac{{{n}_{N{{O}_{3}}^{-}}}-3{{n}_{A{{l}^{3+}}}}}{2}=0,075\ mol\xrightarrow{BT:\,Fe}{{n}_{Fe(\text{r }\!\!{\scriptscriptstyle 3\!/\!{ }_4}\!\!\text{ n)}}}={{n}_{Fe{{(N{{O}_{3}})}_{3}}}}-{{n}_{F{{e}^{2+}}}}=0,075\ mol$

$\Rightarrow {{m}_{\text{r }\!\!{\scriptscriptstyle 3\!/\!{ }_4}\!\!\text{ n}}}=64{{n}_{Cu}}+56{{n}_{Fe}}=0,15.64+0,075.56=13,8\,(g)$