Cho 56,36 gam hỗn hợp X gồm Mg, Fe(NO3)3, FeCl2 và Fe3O4 vào dung dịch chứa 1,82

Đáp án B

$$\begin{gathered} 56,36X\stackrel{HCl:1,82}{\to }\left\{\begin{array}{l}\left\{\begin{array}{l}NO:0,08\\ N_{2}O:0,06\end{array}\right.\\ Y\stackrel{AgN{O}_3}{\to }\left\{\begin{array}{l}NO:0,045\\ 298,31\left\{\begin{array}{l}Ag\\ AgCl\end{array}\right.\end{array}\right.\end{array}\right. \\ {n}_{NO}=0,045\to n_{H^{+}\left(Y\right)}=4n_{NO}=0,18\to n_{HCl\left(Y\right)}=0,18 \end{gathered}$$

BTKL: $m_X+m_{HCl}=m_{\uparrow }+m_{HCl\left(Y\right)}+m_{muoi}+m_{H_{2}O}$

$\to m_{H_{2}O}=13,32\to n_{H_{2}O}=0,74$

BT H: $n_{N{H}_4^{+}\left(Y\right)}=\frac{n_{HCl}-n_{HCl\left(Y\right)}-2n_{H_{2}O}}{4}=0,04$

BT N: $3n_{Fe{\left(N{O}_3\right)}_3}=n_{NO}+2n_{N_{2}O}+n_{N{H}_4^{+}}\to n_{Fe{\left(N{O}_3\right)}_3}=0,08$

BT O: $4n_{F{e}_3{O}_4}+9n_{Fe{\left(N{O}_3\right)}_3}=n_{NO}+n_{N_{2}O}+n_{H_{2}O}\to n_{F{e}_3{O}_4}=0,04$

Gọi $\left\{\begin{array}{l}Mg:x\\ FeC{l}_2:y\\ Ag:z\end{array}\right.$

BT Cl: $n_{AgCl}=2y+1,82$

Bảo toàn e cho cả quá trình: $2\text{x}+y+0,04=0,08.3+0,06.8+0,04.8+z+0,045.3$  (1)

$24\text{x}+127y+232.0,04+0,08.242=56,36$ (2)

$108\text{z}+143,5\left(2y+1,82\right)=298,31$ (3)

$\to \left\{\begin{array}{l}x=0,52\\ y=0,12\\ z=0,025\end{array}\right.$.

$\to \%m_{FeC{l}_2}=27,04\%$