Cho 27,04 gam hỗn hợp rắn X gồm Fe, FeO, Fe3O4, Fe2O3 và Fe(NO3)2

* Hướng dẫn giải

Đáp án C 

$\begin{array}{l}+ Trong X: n_{Fe}=x; n_{FeO}=3y; n_{F{e}_3{O}_4}=2y; n_{F{e}_2{O}_3}=y; n_{Fe{\left(N{O}_3\right)}_2}=z\\ + \left\{\begin{array}{l}PƯ1:Xdd Y+...\\ PƯ2: YNO+...\end{array}\right.\Rightarrow Y chứa \left\{\begin{array}{l}C{l}^{-}: 0,88 mol\\ H^{+}, F{e}^{2+}, F{e}^{3+}\end{array}\right.\\ + \left\{\begin{array}{l}{n}_{H^{+} trong Y}=4n_{NO}=0,08\\ n_{F{e}^{2+} trong Y}=3n_{NO}+n_{Ag}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{H_{2}O tạo ra ở PƯ1}=\frac{0,88+0,04-0,08}{2}=0,42\\ n_{F{e}^{2+} trong Y}=3.0,02+\frac{(133,84-0,88.143,5)}{108}=0,13\end{array}\right.\\ + BT~NT cho Y\Rightarrow n_{F{e}^{3+}}=0,18\Rightarrow m_{các chất \tan trong Y}=48,68\\ + \left\{\begin{array}{l}{n}_{N{O}_2}+n_{N_{2}O}=0,12\\ 46n_{N{O}_2}+44n_{N_{2}O}=m_X+m_{axit}-m_{ chất \tan trong Y}-m_{H_{2}O}=5,44\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{N{O}_2}=0,08\\ n_{N_{2}O}=0,04\end{array}\right.\\ + \left\{\begin{array}{l}BTNT N: n_{Fe{\left(N{O}_3\right)}_2}=(0,08+0,04.2-0,04)/2=0,06\\ BTNT Fe: x+3y+3.2y+2y+0,06=0,31\\ BTE: 3x+3y+2y+0,06=0,08+0,04.8+0,13\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=0,14 (\approx 54\%)\\ y=0,01\end{array}\right.\end{array}$