Đáp án:
\( \% {m_{Al}} = 51,43\% \)
\( \% {m_{A{l_2}{O_3}}} = 48,57\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2Al + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}\)
\(A{l_2}{O_3} + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}O\)
Ta có:
\({n_{{H_2}}} = \frac{{13,44}}{{22,4}} = 0,6{\text{ mol = }}\frac{3}{2}{n_{Al}}\)
\( \to {n_{Al}} = 0,4{\text{ mol}}\)
\( \to {m_{Al}} = 0,4.27 = 10,8{\text{ gam}}\)
\( \to \% {m_{Al}} = \frac{{10,8}}{{21}} = 51,43\% \)
\( \to \% {m_{A{l_2}{O_3}}} = 100\% - 51,43\% = 48,57\% \)