Đốt cháy m gam hỗn hợp Mg, Fe trong oxi một thời gian, thu được (m + 4,16) gam hỗn hợp X chứa các oxit.

* Hướng dẫn giải

$\begin{array}{l}+\underset{m}{\underbrace{\left\{\begin{array}{l}Mg\\ Fe\end{array}\right\}}}\underset{t^o}{\overset{O_2}{\to }}\underset{(m+4,16)}{\underbrace{\left\{oxit\right\}}}\stackrel{HCl}{\to }\underset{(3m+1,82)}{\underbrace{\left\{muối\right\}}}\stackrel{AgN{O}_3}{\to }\underset{(9m+4,06)}{\underbrace{\left\{\begin{array}{l}AgCl\downarrow \\ Ag\downarrow \end{array}\right\}}}\\ +\left\{\begin{array}{l}BTKL:n_{O^{2-}}=\frac{4,16}{16}=0,26\\ BTĐT:n_{C{l}^{-}}=2n_{O^{2-}}=0,52\end{array}\right.\Rightarrow \underset{m_{muối}}{\underbrace{(3m+1,82)}}-\underset{m_{oxit}}{\underbrace{(m+4,16)}}=\underset{m_{C{l}^{-}}}{\underbrace{0,52.35,5}}-\underset{m_{O^{2-}}}{\underbrace{4,16}}\\ \Rightarrow m=8,32.\end{array}$

$$\begin{gathered} \begin{array}{l}+n_{AgCl}=n_{C{l}^{-}}=0,52\Rightarrow n_{Ag}=\frac{9.8,32+4,06-0,52.143,5}{108}=0,04.\\ +\left\{\begin{array}{l}{m}_{(Fe,Mg)}=56n_{Fe}+24n_{Mg}=8,32\\ BTE:3n_{Fe}+2n_{Mg}=2n_{O^{2-}}+n_{Ag}=0,56\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{Fe}=0,08\\ n_{Mg}=0,16\end{array}\right.\\ \Rightarrow TrongX:n_{Fe}:n_{Mg}:n_O=0,08:0,16:0,26=4:8:13\end{array} \\ +\left\{\begin{array}{l}{m}_{Oxit/3,75mgam}=56.4x+24.8x+16.13x=3,75.8,32=31,2\\ m_{muối}=242.4x+148.8x\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=0,05\\ \boxed{m_{muối}=107,6}\end{array}\right. \end{gathered}$$

Chọn A