Lời giải:\(X\left\{ \begin{gathered}
Fe:0,1\,\,mol \hfill \\
Fe{(N{O_3})_2}:0,15\,\,mol \hfill \\
Al \hfill \\
\end{gathered} \right.\xrightarrow[{0,61}]{{ + HCl}}\left| \begin{gathered}
47,455\,\,gam\,\,Y\left\{ \begin{gathered}
\left\{ {F{e^{2 + }};F{e^{3 + }}} \right\}0,25\,\,mol \hfill \\
A{l^{3 + }} \hfill \\
NH_4^ + :0,01\,\,mol \hfill \\
C{l^ - }:0,61\,\,mol \hfill \\
NO_3^ - du :0,17\,\,mol \hfill \\
\end{gathered} \right. \hfill \\
0,105\,\,mol\,\,Z\left\{ \begin{gathered}
NO:0,09 \hfill \\
{N_2}O:0,015 \hfill \\
\end{gathered} \right. \hfill \\
\end{gathered} \right.\)
\({n_{NH_4^ - }} = \frac{{{n_{{H^ + }}} - 4{n_{NO}} - 10{n_{{N_2}O}}}}{{10}} = \frac{{0,61 - 4.0,09 - 10.0,015}}{{10}} = 0,01\,\,mol\)
\(\xrightarrow{{BT.N}}{n_{NO_3^ - (Y)}} = 2{n_{Fe{{(N{O_3})}_2}}} - {n_{NH_4^ + }} - {n_{NO}} - 2{n_{{N_2}O}} = 2.0,15 - 0,01 - 0,09 - 2.0,015 = 0,17\,\,mol\)
\( \to {m_{Al}} = {m_{A{l^{3 + }}(Y)}} = 47,455 - 56.0,25 - 18.0,01 - 35,5.0,61 - 62.0,17 = 1,08\,\,gam\)
