Hòa tan hoàn toàn 8,6 gam hỗn hợp Al, Mg, Fe, Zn vào 100 gam

* Hướng dẫn giải

Đáp án A

$$\begin{gathered} + \mathrm{Từ} \mathrm{giả} \mathrm{thiết} \mathrm{suy} \mathrm{ra} \mathrm{X} \mathrm{chứa} \left\{\begin{array}{l}{\mathrm{MgSO}}_4, {\mathrm{FeSO}}_4, {\mathrm{Fe}}_2({\mathrm{SO}}_4{)}_3\\ {\mathrm{Al}}_2({\mathrm{SO}}_4{)}_3, {\mathrm{ZnSO}}_4, {\mathrm{K}}_2{\mathrm{SO}}_4\end{array}\right. \\ + {\mathrm{n}}_{{\mathrm{KNO}}_3}=\mathrm{x}; {\mathrm{n}}_{{\mathrm{H}}_2{\mathrm{SO}}_4}=2\mathrm{x}\Rightarrow {\mathrm{m}}_{\mathrm{muối}/\mathrm{X}}=8,6+39\mathrm{x}+2\mathrm{x}.96=43,25\Rightarrow \mathrm{x}=0,15. \\ + {\mathrm{n}}_{{\mathrm{H}}_2}=\mathrm{y}\Rightarrow {\mathrm{m}}_{\mathrm{Y}}=50\mathrm{y}, {\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}=0,3-\mathrm{y}. \\ + \left\{\mathrm{BTKL}: {\mathrm{m}}_{\mathrm{kim} \mathrm{loại}}+{\mathrm{m}}_{({\mathrm{KNO}}_3, {\mathrm{H}}_2{\mathrm{SO}}_4)}={\mathrm{m}}_{\mathrm{muối}}+{\mathrm{m}}_{\mathrm{Y}}+{\mathrm{m}}_{{\mathrm{H}}_2\mathrm{O}} \\ \mathrm{BTKL}: {\mathrm{m}}_{\mathrm{kim} \mathrm{loại}}+{\mathrm{m}}_{\mathrm{dd} ({\mathrm{KNO}}_3, {\mathrm{H}}_2{\mathrm{SO}}_4)}={\mathrm{m}}_{\mathrm{dd} \mathrm{X}}+{\mathrm{m}}_{\mathrm{Y}}\right. \\ \Rightarrow \left\{8,6+0,15.101+0,3.98=43,25+50\mathrm{y}+18(0,3-\mathrm{y}) \\ 8,6+100={\mathrm{m}}_{\mathrm{dd} \mathrm{X}}+50\mathrm{y}\right.\Rightarrow \left\{\mathrm{y}=0,140625 \\ {\mathrm{m}}_{\mathrm{dd} \mathrm{X}}=101,56875\right. \\ + \mathrm{KOH}+\mathrm{X}\stackrel{}{\to }\mathrm{dd} \mathrm{chỉ} \mathrm{chứa} {\mathrm{K}}_2{\mathrm{SO}}_4+ \mathrm{Z}\downarrow \Rightarrow \left\{{\mathrm{n}}_{\mathrm{KOH}}=2{\mathrm{n}}_{{\mathrm{K}}_2{\mathrm{SO}}_4}-{\mathrm{n}}_{{\mathrm{KNO}}_3}=0,45 \\ {\mathrm{n}}_{{\mathrm{OH}}^{-} \mathrm{trong} \mathrm{Z}}={\mathrm{n}}_{\mathrm{KOH}}=0,45\right. \\ + \mathrm{Khi} \mathrm{nung} \mathrm{Z} \mathrm{sẽ} \mathrm{xảy} \mathrm{ra} \mathrm{quá} \mathrm{trình}: \left\{0,45 \mathrm{mol} {\mathrm{OH}}^{-}\stackrel{}{\to }0,225 \mathrm{mol} {\mathrm{O}}^{2-} \\ {\mathrm{O}}_2\stackrel{\mathrm{oxi} \mathrm{hóa} {\mathrm{Fe}}^{2+}}{\to } \mathrm{z} \mathrm{mol} {\mathrm{O}}^{2-}\right. \\ \Rightarrow 0,225.16+16\mathrm{z}=12,6-8,6\Rightarrow \mathrm{z}=0,025\Rightarrow \mathrm{BTE}: {\mathrm{n}}_{{\mathrm{Fe}}^{2+}}=2{\mathrm{n}}_{{\mathrm{O}}^{2-}}=0,05 \\ \Rightarrow \mathrm{C}{\%}_{{\mathrm{FeSO}}_4}=\frac{152.0,05}{101,56875}.100\%=7,48\% \mathrm{gần} \mathrm{nhất} \mathrm{với} \boxed{7,5\%} \end{gathered}$$