Hỗn hợp X gồm Al, Fe, Cu. Lấy 3,31 gam X cho vào d...

- Hướng dẫn giải

Phương pháp giải:

Bảo toàn e, bảo toàn khối lượng

Giải chi tiết:

\(\eqalign{ & \underbrace {X\left\{ \matrix{ Al:x \hfill \cr Fe:y \hfill \cr Cu:z \hfill \cr} \right.}_{3,31g} \to \left[ \matrix{ \buildrel { + HCl} \over \longrightarrow \uparrow {H_2}:0,035 \hfill \cr \buildrel { + {H_2}S{O_4}\,du} \over \longrightarrow \,muoi:\,10,51g \hfill \cr} \right. \cr & Al{\rm{ }} + {\rm{ }}3HCl{\rm{ }} \to {\rm{ }}AlC{l_3} + {\rm{ }}1,5{H_2} \uparrow \cr & Fe{\rm{ }} + {\rm{ }}2HCl{\rm{ }} \to {\rm{ }}FeC{l_2} + {\rm{ }}{H_2} \uparrow \cr & 2Al{\rm{ }} + {\rm{ }}6H{}_2S{O_4} \to {\rm{ }}A{l_2}{(S{O_4})_3} + {\rm{ }}3S{O_2} \uparrow {\rm{ }} + {\rm{ }}6{H_2}O \cr & 2Fe{\rm{ }} + {\rm{ }}6{H_2}S{O_4} \to {\rm{ }}F{e_2}{(S{O_4})_3} + {\rm{ }}3S{O_2} \uparrow {\rm{ }} + {\rm{ }}6{H_2}O \cr & Cu{\rm{ }} + {\rm{ }}2{H_2}S{O_4} \to {\rm{ }}CuS{O_4} + {\rm{ }}S{O_2} \uparrow {\rm{ }} + {\rm{ }}{H_2}O \cr & \left\{ \matrix{ 27x + 56y + 64z = 3,31 \hfill \cr \buildrel {{H_2}:0,035} \over \longrightarrow 1,5x + y = 0,035 \hfill \cr \buildrel {muoi = 10,51g} \over \longrightarrow 342\,.\,0,5x + 400\,.\,0,5y + 160z = 10,51 \hfill \cr} \right. \to \left\{ \matrix{ x = 0,01 \hfill \cr y = 0,02 \hfill \cr z = 0,03 \hfill \cr} \right. \to \left\{ \matrix{ \% Al = 8,16\% \hfill \cr \% Fe = 33,84\% \hfill \cr \% Cu = 58\% \hfill \cr} \right. \cr} \)