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

* Hướng dẫn giải

$\begin{array}{l}+\underset{\mathrm{39,36}gam}{\underbrace{\left\{\begin{array}{l}FeO\\ F{e}_3{O}_4\end{array}\right\}}}\underset{F{e}_3{O}_4=FeO+F{e}_2{O}_3}{\overset{quy\text{\hspace{0.17em}đổi}}{\to }}\underset{\mathrm{39,36}gam}{\underbrace{\left\{\begin{array}{l}FeO:xmol\\ F{e}_2{O}_3:ymol\end{array}\right\}}}\stackrel{H_{2}S{O}_{4}loãngdư}{\to }\left\{\begin{array}{l}FeS{O}_4:xmol\\ F{e}_2{\left(S{O}_4\right)}_3:ymol\\ H_{2}S{O}_4\end{array}\right\}\stackrel{KMn{O}_4}{\to }\left\{\begin{array}{l}MnS{O}_4\\ F{e}_2{\left(S{O}_4\right)}_3\\ K_{2}S{O}_4\end{array}\right\}\\ +\left\{\begin{array}{l}{m}_{hỗnhợp}=\mathrm{39,36}\\ BTE:n_{FeS{O}_4}=5n_{KMn{O}_4}\end{array}\right.\Rightarrow \left\{\begin{array}{l}72x+160y=\mathrm{39,36}\\ x=\mathrm{0,28}\end{array}\right.\Rightarrow \left\{\begin{array}{l}y=\mathrm{0,12}\\ BTE:n_{Cu}=n_{F{e}_2{\left(S{O}_4\right)}_3}=\mathrm{0,12}\end{array}\right.\Rightarrow \boxed{m_{Cu}=\mathrm{7,68}gam}\end{array}$