Thực hiện phản ứng nhiệt nhôm hỗn hợp X gồm Al và Fe2O3

* Hướng dẫn giải

Đáp án A

$$\begin{gathered} + \mathrm{P}1: \left\{{\mathrm{n}}_{\mathrm{Al}}=\frac{2{\mathrm{n}}_{{\mathrm{H}}_2}}{3}=\frac{2.1,008}{3.22,4}=0,03 \\ {\mathrm{n}}_{\mathrm{Fe}}=\frac{3,36}{56}=0,06\right.\Rightarrow \left\{{\mathrm{n}}_{{\mathrm{Al}}_2{\mathrm{O}}_3}={\mathrm{n}}_{{\mathrm{Fe}}_2{\mathrm{O}}_3/\mathrm{X}}=0,5{\mathrm{n}}_{\mathrm{Fe}}=0,03 \\ {\mathrm{n}}_{\mathrm{Fe}}:{\mathrm{n}}_{\mathrm{Al}}:{\mathrm{n}}_{{\mathrm{Al}}_2{\mathrm{O}}_3}=2:1:1 \\ {\mathrm{m}}_{\mathrm{P}1}=7,23\right. \\ + \mathrm{P}2: \left\{{\mathrm{n}}_{\mathrm{Fe}}=2\mathrm{x}; {\mathrm{n}}_{\mathrm{Al}}=\mathrm{x}; {\mathrm{n}}_{{\mathrm{Al}}_2{\mathrm{O}}_3}=\mathrm{x} \\ 2\mathrm{x}.56+27\mathrm{x}+102\mathrm{x}=28,92-7,23=21,69\right.\Rightarrow \mathrm{x}=0,09 \\ + \underset{\mathrm{P}2}{\underbrace{\left\{\mathrm{Fe}: 0,18 \\ \mathrm{Al}: 0,09 \\ {\mathrm{Al}}_2{\mathrm{O}}_3: 0,09\right\}}}\underset{1,52 \mathrm{mol}}{\overset{{\mathrm{HNO}}_3 (\mathrm{v} \mathrm{ừa} \mathrm{đủ})}{\to }}\underset{\mathrm{Z}}{\underbrace{\left\{{\mathrm{Fe}}^{3+}, {\mathrm{Fe}}^{2+} \\ {\mathrm{Al}}^{3+}, {{\mathrm{NO}}_3}^{-} \\ {{\mathrm{NH}}_4}^{+}: \mathrm{a}\right\}}}+\underset{0,17}{\underbrace{\mathrm{NO}\uparrow }}+{\mathrm{H}}_2\mathrm{O} \\ + \left\{\mathrm{BTNT} \mathrm{H}: {\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}=\frac{1,52-4\mathrm{a}}{2}=0,76-2\mathrm{a} \\ \mathrm{BTNT} \mathrm{N}: {\mathrm{n}}_{{{\mathrm{NO}}_3}^{-}/\mathrm{Z}}=1,52-0,17-\mathrm{a}=1,35-\mathrm{a} \\ \mathrm{BTNT} \mathrm{O}: 0,09.3+1,52.3=3(1,35-\mathrm{a})+0,17+(0,76-2\mathrm{a})\right.\Rightarrow \mathrm{a}=0,03 \\ \Rightarrow {\mathrm{m}}_{\mathrm{muối} \mathrm{Z}}={\mathrm{m}}_{\mathrm{ion} \mathrm{kim} \mathrm{loại}}+{\mathrm{m}}_{{{\mathrm{NH}}_4}^{+}}+{\mathrm{m}}_{{{\mathrm{NO}}_3}^{-}}=99,75 \boxed{\mathrm{gần} \mathrm{nhất} \mathrm{với} 100} \end{gathered}$$