Hòa tan hết hỗn hợp gồm Mg, Fe3O4 và Fe(NO3)2 trong dung dịch chứa 1,44

* Hướng dẫn giải

Đáp án A

$\left\{\begin{array}{l}NO:a\\ H_2:b\end{array}\right.\to \left\{\begin{array}{l}a+b=0,16\\ 30\text{a}+2b=11,\mathrm{5.2.0},16\end{array}\right.\to \left\{\begin{array}{l}a=0,12\\ b=0,04\end{array}\right.$

$X\left\{\begin{array}{l}F{\text{e}}^{3+}:x\\ F{\text{e}}^{2+}:y\\ M{g}^{2+}:z\\ N{H}_4^{+}:t\\ C{l}^{-}:1,44\end{array}\right.\stackrel{NaOH}{\to }47,24\left\{\begin{array}{l}F\text{e}{\left(OH\right)}_3:x\\ F\text{e}{\left(OH\right)}_2:y\\ Mg{\left(OH\right)}_2:z\end{array}\right.\to 107\text{x}+90y+58\text{z}=47,24$(1)

BTĐT: $3\text{x}+2y+2\text{z}+t=1,44$        (2)

BT N: $n_{F\text{e}{\left(N{O}_3\right)}_2}=\frac{t+0,12}{2}=0,5t+0,06$

BT Fe: $n_{F{\text{e}}_3{O}_4}=\frac{x+y-0,5t-0,06}{3}$

$$\begin{gathered} n_{H^{+}}=8n_{F{\text{e}}_3{O}_4}+4n_{NO}+2n_{H_2}+10n_{N{H}_4^{+}} \\ \to 8\frac{x+y-0,5t-0,06}{3}+0,12.4+0,04.2+10t=1,44 \\ {m}_X=75,28\to 56\left(x+y\right)+24\text{z}+18t+1,44.35,5=75,28 \\ \to \left\{\begin{array}{l}x=0,14\\ y=0,12\\ z=0,37\\ t=0,04\end{array}\right.\to n_{Ag}=n_{F{\text{e}}^{2+}}=0,12 \end{gathered}$$

$\to m_{\downarrow }=m_{Ag}+m_{AgCl}=0,12.108+1,44.143,5=219,6$ gam