Đáp án:
\(\% {m_{Al}} = 45,76\% ;\% {m_{Cu}} = 54,24\% \)
m=4,9 gam ; a=24,5
Giải thích các bước giải:
Sơ đồ phản ứng:
\(\left\{ \begin{gathered}
Al \hfill \\
Cu \hfill \\
\end{gathered} \right.\xrightarrow{{{H_2}S{O_4}}}\left[ \begin{gathered}
\left\{ \begin{gathered}
A{l_2}{(S{O_4})_3} \hfill \\
CuS{O_4} \hfill \\
\end{gathered} \right.\xrightarrow{{ + NaOH}}Cu{(OH)_2}\xrightarrow{{}}CuO \hfill \\
{H_2}S;S{O_2} \hfill \\
\end{gathered} \right.\)
Ta có:
\({m_{CuO}} = 4{\text{gam}} \to {{\text{n}}_{CuO}} = {n_{Cu}} = \frac{4}{{80}} = 0,05{\text{ mol}} \to {{\text{m}}_{Cu}} = 0,05.64 = 3,2 \to {m_{Al}} = 2,7 \to {n_{Al}} = 0,1\)
Gọi số mol \({H_2}S;S{O_2}\) lần lượt là x, y.
\( \to x + y = \frac{{1,792}}{{22,4}} = 0,08{\text{ mol}}\)
Bảo toàn e:
\(3{n_{Al}} + 2{n_{Cu}} = 8{n_{{H_2}S}} + 2{n_{S{O_2}}} \to 8x + 2y = 0,1.3 + 0,05.2 = 0,4\)
Giải được: x=y=0,04
\( \to {M_{hh}} = \frac{{0,04.34 + 0,04.64}}{{0,08}} = 49 \to a = \frac{{49}}{2} = 24,5\)
\(m = {m_{Cu{{(OH)}_2}}} = 0,05.(64 + 17.2) = 4,9{\text{ gam}}\)
\(\% {m_{Al}} = \frac{{2,7}}{{5,9}} = 45,76\% \to \% {m_{Cu}} = 54,24\% \)
Ta có:
\({n_{A{l_2}{{(S{O_4})}_3}}} = \frac{1}{2}{n_{Al}} = 0,05{\text{ mol; }}{{\text{n}}_{CuS{O_4}}} = {n_{Cu}} = 0,05{\text{ mol}}\)
\( \to {m_{A{l_2}{{(S{O_4})}_3}}} = 0,05.(27.2 + 96.3) = 17,1{\text{gam; }}{{\text{m}}_{CuS{O_4}}} = 0,05.160 = 8{\text{gam}}\)
Bảo toàn S:
\({n_{{H_2}S{O_4}}} = 3{n_{A{l_2}{{(S{O_4})}_3}}} + {n_{CuS{O_4}}} + {n_{{H_2}S}} + {n_{S{O_2}}} = 0,05.3 + 0,05 + 0,04 + 0,04 = 0,28{\text{mol}} \to {{\text{m}}_{{H_2}S{O_4}}} = 0,28.98 = 27,44{\text{ gam}}\)
