1)

Phản ứng xảy ra:

\(2Al + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}\)

Ta có: 

\({n_{Al}} = \frac{{13,5}}{{27}} = 0,5{\text{ }}mol \to {n_{{H_2}}} = \frac{3}{2}{n_{Al}} = 0,75{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,75.22,4 = 16,8{\text{ lít}}\)

\({n_{HCl}} = 3{n_{Al}} = 1,5{\text{ mol}} \to {{\text{m}}_{HCl}} = 1,5.36,5 = 54,75{\text{ gam}} \to {\text{m = }}\frac{{54,75}}{{12\% }} = 456,25{\text{ gam}}\)

\(F{e_2}{O_3} + 3{H_2}\xrightarrow{{{t^o}}}2Fe + 3{H_2}O\)

Ta có: 

\({n_{F{e_2}{O_3}}} = \frac{{16}}{{56.2 + 16.3}} = 0,1{\text{ mol < }}\frac{1}{3}{n_{{H_2}}} \to {n_{Fe}} = 2{n_{F{e_2}{O_3}}} = 0,2{\text{ mol}} \to {{\text{m}}_{Fe}} = 0,2.56 = 11,2{\text{ gam}}\)

Câu 2:

Phản ứng xảy ra:

\(2Al + 3{H_2}S{O_4}\xrightarrow{{}}A{l_2}{(S{O_4})_3} + 3{H_2}\)

Ta có: 

\({n_{Al}} = \frac{{2,16}}{{27}} = 0,08{\text{ mol;}}{{\text{n}}_{{H_2}S{O_4}}} = 0,2.1 = 0,2{\text{ mol > }}\frac{3}{2}{n_{Al}}\)

Vậy axit dư.

\( \to {n_{{H_2}}} = \frac{3}{2}{n_{Al}} = 0,12{\text{ mol}} \to {V_{{H_2}}} = 0,12.22,4 = 2,688{\text{ lít}}\)

Dung dịch thu được chứa \({n_{A{l_2}{{(S{O_4})}_3}}} = \frac{1}{2}{n_{Al}} = 0,04{\text{ mol; }}{{\text{n}}_{{H_2}S{O_4}{\text{ dư}}}} = 0,2 - 0,04.3 = 0,08{\text{ mol}} \to {{\text{C}}_{M{\text{ A}}{{\text{l}}_2}{{(S{O_4})}_3}}} = \frac{{0,04}}{{0,2}} = 0,2M;{C_{M{\text{ }}{{\text{H}}_2}S{O_4}}} = \frac{{0,08}}{{0,2}} = 0,4M\)

3)

Phản ứng xảy ra:

\(2Al + 3{H_2}S{O_4}\xrightarrow{{}}A{l_2}{(S{O_4})_3} + 3{H_2}\)

Ta có: 

\({n_{Al}} = \frac{{2,16}}{{27}} = 0,08{\text{ mol; }}{{\text{m}}_{{H_2}S{O_4}}} = 19,6.10\%  = 1,96{\text{ gam}} \to {{\text{n}}_{{H_2}S{O_4}}} = \frac{{1,96}}{{98}} = 0,02{\text{ mol < }}\frac{3}{2}{n_{Al}}\) vậy Al dư.

\( \to {n_{{H_2}}} = {n_{{H_2}S{O_4}}} = 0,02{\text{ mol}} \to {\text{V = 0}}{\text{,02}}{\text{.22}}{\text{,4 = 0}}{\text{,448 lít}}\)

\({n_{Al{\text{phản ứng}}}} = \frac{2}{3}{n_{{H_2}S{O_4}}} = \frac{{0,04}}{3} \to {n_{A{l_2}{{(S{O_4})}_3}}} = \frac{{0,02}}{3} \to {m_{A{l_2}{{(S{O_4})}_3}}} = \frac{{0,02}}{3}.(27.2 + 96.3) = 2,28{\text{ gam}}\)

BTKL: 

\({m_{dd{\text{sau phản ứng}}}} = \frac{{0,04}}{3}.27 + 19,6 - 0,02.2 = 19,92{\text{ gam}} \to {\text{C}}{{\text{\% }}_{A{l_2}{{(S{O_4})}_3}}} = \frac{{2,28}}{{19,92}} = 11,45\% \)