Đáp án:
\({m_{A{l_2}{O_3}{\text{ dư}}}} = 0,34{\text{ gam}}\)
\( {m_{AlC{l_3}}} = 4,45{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(A{l_2}{O_3} + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}O\)
Ta có:
\({n_{A{l_2}{O_3}}} = \frac{{2,04}}{{27.2 + 16.3}} = 0,02{\text{ mol}}\)
\({m_{HCl}} = 50.7,3\% = 3,65{\text{ gam}} \to {{\text{n}}_{HCl}} = \frac{{3,65}}{{36,5}} = 0,1{\text{ mol}}\)
\({n_{HCl}} < 6{n_{A{l_2}{O_3}}}\) nên \(Al_2O_3\) dư
\( \to {n_{A{l_2}{O_3}{\text{ dư}}}} = 0,02 - \frac{{0,1}}{6} = \frac{1}{{300}}{\text{ mol}}\)
\( \to {m_{A{l_2}{O_3}{\text{ dư}}}} = \frac{1}{{300}}.(27.2 + 16.3) = 0,34{\text{ gam}}\)
\( \to {n_{AlC{l_3}}} = \frac{1}{3}{n_{HCl}} = \frac{{0,1}}{3}{\text{ mol}}\)
\( \to {m_{AlC{l_3}}} = \frac{{0,1}}{3}.(27 + 35,5.3) = 4,45{\text{ gam}}\)
