Đáp án:
\(\% {m_{Al}} = 5,028\% ; \% {m_{A{l_2}{O_3}}} = 94,972\% \)
\({m_{AlC{l_3}}} = 258,456{\text{ gam}}\)
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:
\({n_{{H_2}}} = \frac{{6,72}}{{22,4}} = 0,3{\text{ mol}} \to {{\text{n}}_{Al}} = \frac{2}{3}{n_{{H_2}}} = 0,2{\text{ mol}}\)
\( \to {m_{Al}} = 0,2.27 = 5,4{\text{ gam}} \to {{\text{m}}_{A{l_2}{O_3}}} = 107,4 - 5,4 = 102{\text{ gam}}\)
\( \to \% {m_{Al}} = \frac{{5,4}}{{107,4}} = 5,028\% \to \% {m_{A{l_2}{O_3}}} = 94,972\% \)
Ta có:
\({n_{A{l_2}{O_3}}} = \frac{{102}}{{27.2 + 16.3}} = 1{\text{ mol}}\)
\( \to {n_{AlC{l_3}}} = {n_{Al}} + 2{n_{A{L_2}{O_3}}} = 1.2 + 2 = 2,2{\text{ mol}}\)
Vì hiệu suất là 88%.
\( \to {n_{AlC{l_3}}} = 2,2.88\% = 1,936{\text{ mol}}\)
\( \to {m_{AlC{l_3}}} = 1,936.(27 + 35,5.3) = 258,456{\text{ gam}}\)
