Nung bột Fe2O3 với a gam bột...

B

- Hướng dẫn giải

\(\begin{gathered} \,\left\{ \begin{gathered} F{e_2}{O_3}:\,\,x\,\,mol \hfill \\ Al \hfill \\ \end{gathered} \right\}\xrightarrow{{{t^o}}}\underbrace {\left\{ \begin{gathered} Fe:\,\,2x\,\,mol \hfill \\ A{l_2}{O_3}:\,\,x\,\,mol \hfill \\ Al\,\,du:\,\,y \hfill \\ \end{gathered} \right\}}_{{m_X} = 11,78\,\,gam}\xrightarrow{{NaOH}}NaAl{O_2} + \underbrace {{H_2} \uparrow }_{0,06\,\,mol} \hfill \\ + \,\,\left\{ \begin{gathered} {m_X} = 11,78 \hfill \\ BTE:\,\,3{n_{Al\,\,du}} = 2{n_{{H_2}}} \hfill \\ \end{gathered} \right. \Rightarrow \left\{ \begin{gathered} 2x.56 + 102x + 27y = 11,78 \hfill \\ 3y = 2.0,06 \hfill \\ \end{gathered} \right. \Rightarrow \left\{ \begin{gathered} y = 0,04 \hfill \\ x = 0,05 \hfill \\ \end{gathered} \right. \hfill \\ \Rightarrow {n_{Al\,\,ban\,\,dau}} = 2x + y = 0,14 \Rightarrow \boxed{{m_{Al}} = 3,78\,\,gam} \hfill \\ \end{gathered} \)