Hỗn hợp X gồm Al và Fe. Cho X tan vừa hết trong 352 ml

* Hướng dẫn giải

$$\begin{gathered} \mathit{Đ}\mathit{á}\mathit{p}\mathbf{ }\mathit{á}\mathit{n}\mathbf{ }\mathit{B} \\ \left\{\begin{array}{l}Al\\ Fe\end{array}\underset{0,88}{\overset{HN{O}_3}{\to }}\left\{\begin{array}{l}\left\{\begin{array}{l}F{e}^{2+}, F{e}^{3+}\\ A{l}^{3+}, N{O_3}^{-}\\ N{H_4}^{+}, x mol\end{array}\right.\\ \left\{\begin{array}{l}NO\\ N_{2}O\end{array}+H_{2}O\right.\end{array}\right.\right. \\ \left\{\begin{array}{l}{n}_{NO}+n_{N_{2}O}=0,1\\ 3n_{NO}+44n_{N_{2}O}=17,1.2.0,1=3,42\end{array}\right. \\ \Rightarrow \left\{\begin{array}{l}{n}_{NO}=0,07\\ n_{N_{2}O}=0,03\end{array}\right. \\ BTNT H:n_{H_{2}O}=\frac{0,88-4x}{2} \\ =0,44-2x \\ BTNT N:n_{N{O_3}^{-}\left(Y\right)}=0,81-0,3.2-x \\ =0,75-x \\ BTNT H: 3.0,88=3\left(0,75-x\right) \\ +0,03+0,07+\left(0,44-2x\right) \\ {m}_{\left(Al,Fe\right)}=m_{muối}+m_{khí}+m_{H_{2}O}-m_{HN{O}_3} \\ \Rightarrow \left\{\begin{array}{l}x=0,03\\ m_{\left(Al,Fe\right)}=8,22\end{array}\right. \\ Y+dd N{H}_3\to kết tủa \\ {n}_{O{H}^{-}}=3n_{A{l}^{3+}}+3n_{F{e}^{3+}}+2n_{F{e}^{2+}} \\ =\left(0,75-0,03\right)-0,03=0,69 \\ {m}_{KT}=m_{KL}+m_{O{H}^{-}} \\ =8,22+0,69.17=19,95 \end{gathered}$$