Đáp án:
a) \(\% {m_{A{l_2}{O_3}}} = 85\% ;\% {m_{Al{\text{ dư}}}} = 15\% \)
b) \({{\text{V}}_{HCl}} = 0,15{\text{ lít}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(4Al + 3{O_2}\xrightarrow{{}}2A{l_2}{O_3}\)
Ta có: \({n_{Al}} = \frac{{5,4}}{{27}} = 0,2{\text{ mol; }}{{\text{n}}_{{O_2}}} = \frac{{3,6}}{{32}} = 0,1125{\text{ mol}} \to {{\text{n}}_{Al}} > \frac{4}{3}{n_{{O_2}}}\) nên Al dư.
\(\to {n_{A{l_2}{O_3}}} = \frac{2}{3}{n_{{O_2}}} = 0,075{\text{ mol; }}{{\text{n}}_{Al{\text{ dư}}}} = 0,2 - \frac{4}{3}{n_{{O_2}}} = 0,05{\text{ mol}}\)
\(\to {m_{A{l_2}{O_3}}} = 0,075.(27.2 + 16.3) = 7,65{\text{ gam; }}{{\text{m}}_{Al{\text{ dư}}}} = 0,05.27 = 1,35{\text{ gam}}\)
\(\to \% {m_{A{l_2}{O_3}}} = \frac{{7,65}}{{7,65 + 1,35}} = 85\% \to \% {m_{Al{\text{ dư}}}} = 15\% \)
Hòa tan A vào HCl
\(A{l_2}{O_3} + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}O\)
\(2Al + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}\)
\(\to {n_{HCl}} = 6{n_{A{l_2}{O_3}}} + 3{n_{Al}} = 0,075.6 + 0,05.3 = 0,6{\text{ mol}} \to {{\text{V}}_{HCl}} = \frac{{0,6}}{4} = 0,15{\text{ lít}}\)
