Đáp án:
\(\% {m_{Al}} = 21,43\% ;\% {m_{Cu}} = 38,1\% ; \% {m_{A{l_2}{O_3}}} = 40,47\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2Al + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}\)
\(A{l_2}{O_3} + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}O\)
Rắn \(D\) là \(Cu\)
\(2Cu + {O_2}\xrightarrow{{{t^o}}}2CuO\)
Ta có:
\({n_{{H_2}}} = \frac{{6,72}}{{22,4}} = 0,3{\text{ mol = }}\frac{3}{2}{n_{Al}} \to {n_{Al}} = 0,2{\text{ mol}}\)
\( \to {m_{Al}} = 0,2.27 = 5,4{\text{ gam}}\)
\({n_{CuO}} = \frac{{12}}{{64 + 16}} = 0,15{\text{ mol = }}{{\text{n}}_{Cu}}\)
\( \to {m_{Cu}} = 0,15.64 = 9,6{\text{ gam}}\)
\( \to \% {m_{Al}} = \frac{{5,4}}{{25,2}} = 21,43\% ;\% {m_{Cu}} = \frac{{9,6}}{{25,2}} = 38,1\% \to \% {m_{A{l_2}{O_3}}} = 40,47\% \)