Hòa tan 17,32 gam hỗn hợp X gồm Mg, Fe3O4 và Fe(NO3)2 cần vừa đúng dung dịch hỗn hợp gồm 1,04 mol HCl và

* Hướng dẫn giải

Chọn đáp án  A. 150,32.

Trong không khí, một khí không màu hóa nâu trong không khí là NO,

$$\begin{gathered} {\overline{\text{M}}}_Z=\mathrm{10,8.2}=\mathrm{21,6}<M_{NO}\to M_{\text{khí còn lại }}<\mathrm{21,6}\to khí còn lại là {\text{H}}_2 \\ Z\left\{\begin{array}{l}NO:a\text{\hspace{0.33em}mol}\\ H_2:b\text{\hspace{0.33em}mol}\end{array}\right.\to \left\{\begin{array}{l}a+b=\frac{\mathrm{2,24}}{\mathrm{22,4}}=\mathrm{0,1}\text{\hspace{0.33em}mol}\\ \text{30a + 2b = 0,1.21,6}\end{array}\right.\to \left\{\begin{array}{l}a=\mathrm{0,07}\\ b=\mathrm{0,03}\end{array}\right. \end{gathered}$$

$X\left\{\begin{array}{l}Mg:x\text{\hspace{0.33em}mol}\\ {\text{Fe}}_3{O}_4:y\text{\hspace{0.33em}mol}\\ {\text{Fe(NO}}_3^{}{)}_2:z\text{\hspace{0.33em}mol}\end{array}\right.\to 24x+232y+180z=\mathrm{17,32}\text{\hspace{0.33em}gam (1)}$

$\begin{array}{l}\stackrel{BTNT.N}{\to }{n}_{N{H}_4^{+}\left(Y\right)}=n_{HN{O}_3}+2n_{Fe{\left(N{O}_3\right)}_2}-n_{NO}=\mathrm{0,08}+2z-\mathrm{0,07}=2z+\mathrm{0,01}\\ {\text{n}}_{H^{+}}=2n_{O\left(F{e}_3{O}_4\right)}+4n_{NO}+10n_{N{H}_4^{+}}+2n_{H_2}\\ \to (\mathrm{1,04}+\mathrm{0,08})=8y+\mathrm{4.0,07}+10(2z+\mathrm{0,01})+\mathrm{2.0,03}=\text{\hspace{0.33em} (2)}\end{array}$

$\text{20,8\hspace{0.33em}gam chất rắn }\left\{\begin{array}{l}MgO\stackrel{BTNT.Mg}{\to }{n}_{MgO}=x\text{\hspace{0.33em}mol}\\ F{e}_2{O}_3\stackrel{BTNT.Fe}{\to }{n}_{F{e}_2{O}_3}=(\mathrm{1,5}y+\mathrm{0,5}z)mol\end{array}\right.$

$\text{Từ (1), (2) và (3) }\to \left\{\begin{array}{l}x=\mathrm{0,4}\\ y=\mathrm{0,01}\\ z=\mathrm{0,03}\end{array}\right.\to Y\left\{\begin{array}{l}M{g}^{2+}:\mathrm{0,4}\text{\hspace{0.33em}mol}\\ {\text{Fe}}^{2+}:a\text{\hspace{0.33em}mol}\\ {\text{Fe}}^{3+}:b\text{\hspace{0.33em}mol}\\ {\text{NH}}_4^{+}:\mathrm{0,07}\text{\hspace{0.33em}mol}\\ {\text{Cl}}^{-}:\mathrm{1,04}\text{\hspace{0.33em}mol}\end{array}\right.$

$\to \left\{\begin{array}{l}\stackrel{BTNT.Fe}{\to }a+b=\mathrm{3.0,01}+\mathrm{0,03}\\ \stackrel{\text{BTĐT}}{\to }2a+3b+\mathrm{2.0,4}+\mathrm{0,07}=\mathrm{1,04}\end{array}\right.\to \left\{\begin{array}{l}a=\mathrm{0,01}\\ b=\mathrm{0,05}\end{array}\right.$

$\to m\text{\hspace{0.33em}gam kết tủa }\left\{\begin{array}{l}Ag\stackrel{BTE}{\to }{n}_{Ag}=n_{F{e}^{2+}}=\mathrm{0,01}\text{\hspace{0.33em}mol}\\ AgCl\stackrel{BTNT.Cl}{\to }{n}_{AgCl}=n_{C{l}^{-}\left(Y\right)}=\mathrm{1,04}\text{\hspace{0.33em}mol}\end{array}\right.$

$\to m=\mathrm{108.0,01}+\mathrm{143,5.1,04}=\mathrm{150,32}\text{\hspace{0.33em}gam -> Đ}áp á\text{n A}$