Hoà tan 39,36 gam hỗn hợp FeO và Fe3O4 vào dung dịch H2SO4 loãng dư

* Hướng dẫn giải

Chọn đáp án A

$$\begin{gathered} + \underset{39,36 gam}{\underbrace{\left\{\begin{array}{l}FeO\\ F{e}_3{O}_4\end{array}\right\}}}\stackrel{F{e}_3{O}_4=FeO+F{e}_2{O}_3}{\to }\underset{39,36 gam}{\underbrace{\left\{\begin{array}{l}FeO: x mol\\ F{e}_2{O}_3: y mol\end{array}\right\}}}\stackrel{KMn{O}_4}{\to }\left\{\begin{array}{l}MnS{O}_4\\ F{e}_2(S{O}_4{)}_3\\ K_{2}S{O}_4\end{array}\right\} \\ + \left\{\begin{array}{l}{m}_{hỗn hợp}=39,36\\ BTE: n_{FeS{O}_4}=5n_{KMn{O}_4}\end{array}\right.\Rightarrow \left\{\begin{array}{l}72x+160y=39,36\\ x=0,28\end{array}\right. \\ \Rightarrow \left\{\begin{array}{l}y=0,12\\ BTE: n_{Cu}=n_{F{e}_2{\left(S{O}_4\right)}_3}=0,12\end{array}\right.\Rightarrow m_{Cu}=7,68 gam \end{gathered}$$