Hòa tan hết hỗn hợp X gồm 5,6 gam Fe; 27 gam Fe(NO3)2

* Hướng dẫn giải

$$\begin{gathered} \mathit{Đ}\mathit{á}\mathit{p}\mathbf{ }\mathit{á}\mathit{n}\mathbf{ }\mathit{A} \\ \left\{\begin{array}{l}{n}_{NO}+n_{N_{2}O}=0,105\\ 30n_{NO}+44n_{N_{2}O}=3,36\end{array}\right. \\ \Rightarrow \left\{\begin{array}{l}{n}_{NO}=0,09\\ n_{N_{2}O}=0,015\end{array}\right. \\ \left\{\begin{array}{l}Fe:0,1 mol\\ Fe{\left(N{O}_3\right)}_2:0,15 mol\\ Al: x mol\end{array} +HCl\right. \\ \to \left\{\begin{array}{l}F{e}^{2+}, F{e}^{3+}\\ A{l}^{3+}\\ N{H_4}^{+}: y\\ N{O_3}^{-}: z\end{array}+\left\{\begin{array}{l}\left\{\begin{array}{l}NO:0,09\\ N_{2}O:0,015\end{array}\right.\\ H_{2}O\end{array}\right.\right. \\ BTNT H: \\ {n}_{H_{2}O}=\frac{0,61-4y}{2}=0,305-2y \\ BTNT O: \\ 0,09+0,015+\left(0,305-2y\right)+3z=0,9 \\ BTNT N: \\ y+z+0,09+0,015.2=0,3 \\ BTKL: \\ 27x+18y+62z+0,25.56+0,61.35,5 \\ =47,455 \\ \Rightarrow \left\{\begin{array}{l}-2y+3z=0,49\\ y+z=0,18\\ 27x+18y+62z=11,8\end{array}\right. \\ \Rightarrow \left\{\begin{array}{l}x=0,04\\ y=0,01\\ z=0,17\end{array}\Rightarrow m_{Al}=1,08\right. \end{gathered}$$