Hòa tan hoàn toàn 8,6 gam hỗn hợp Al, Mg, Fe, Zn v...

A

- Hướng dẫn giải

\(X\,\,\left\{ \begin{gathered} MgS{O_4},\,\,FeS{O_4},\,\,F{e_2}{(S{O_4})_3} \hfill \\ A{l_2}{(S{O_4})_3},\,\,ZnS{O_4},\,\,{K_2}S{O_4} \hfill \\ \end{gathered} \right.\)

\(\begin{gathered} + \,\,{n_{KN{O_3}}} = x;\,\,{n_{{H_2}S{O_4}}} = 2x \Rightarrow {m_{muoi/X}} = 8,6 + 39x + 2x.96 = 43,25 \Rightarrow x = 0,15. \hfill \\ + \,\,{n_{{H_2}}} = y \Rightarrow {m_Y} = 50y,\,\,{n_{{H_2}O}} = 0,3 - y. \hfill \\ + \,\,\left\{ \begin{gathered} BTKL:\,\,{m_{kl}} + {m_{(KN{O_3},\,\,{H_2}S{O_4})}} = {m_{muoi}} + {m_Y} + {m_{{H_2}O}} \hfill \\ BTKL:\,\,{m_{kl}} + {m_{dd\,\,(KN{O_3},\,\,{H_2}S{O_4})}} = {m_{dd\,\,X}} + {m_Y} \hfill \\ \end{gathered} \right. \hfill \\ \Rightarrow \left\{ \begin{gathered} 8,6 + 0,15.101 + 0,3.98 = 43,25 + 50y + 18(0,3 - y) \hfill \\ 8,6 + 100 = {m_{dd\,\,X}} + 50y \hfill \\ \end{gathered} \right. \Rightarrow \left\{ \begin{gathered} y = 0,140625 \hfill \\ {m_{dd\,\,X}} = 101,56875 \hfill \\ \end{gathered} \right. \hfill \\ + \,\,KOH + X\xrightarrow{{}}dd\,\,{K_2}S{O_4} + \,\,Z \downarrow \Rightarrow \left\{ \begin{gathered} {n_{KOH}} = 2{n_{{K_2}S{O_4}}} - {n_{KN{O_3}}} = 0,45 \hfill \\ {n_{O{H^ - }\,\,trong\,\,Z}} = {n_{KOH}} = 0,45 \hfill \\ \end{gathered} \right. \hfill \\ + \,\,Khi\,\,nung\,\,Z\left\{ \begin{gathered} 0,45\,\,mol\,\,O{H^ - }\xrightarrow{{}}0,225\,\,mol\,\,{O^{2 - }} \hfill \\ {O_2}\xrightarrow{{oxh\,F{e^{2 + }}}}\,\,z\,\,mol\,\,{O^{2 - }} \hfill \\ \end{gathered} \right. \hfill \\ \Rightarrow 0,225.16 + 16z = 12,6 - 8,6 \Rightarrow z = 0,025 \Rightarrow BTE:\,\,{n_{F{e^{2 + }}}} = 2{n_{{O^{2 - }}}} = 0,05 \hfill \\ \Rightarrow C{\% _{FeS{O_4}}} = \frac{{152.0,05}}{{101,56875}}.100\% = 7,48\% \,\, \hfill \\ \end{gathered} \)