Đáp án:

\({V_{{H_2}}} = 30,24{\text{ lít}}\)

\( C{\% _{HCl}}= 49,275\% \)

\(C{\% _{AlC{l_3}}} = 54,67\% \)

Giải thích các bước giải:

Phản ứng xảy ra:

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

Ta có:

\({n_{Al}} = \frac{{24,3}}{{27}} = 0,9{\text{ mol}}\)

\( \to {n_{{H_2}}} = \frac{3}{2}{n_{Al}} = \frac{3}{2}.0,9 = 1,35{\text{ mol}}\)

\( \to {V_{{H_2}}} = 1,35.22,4 = 30,24{\text{ lít}}\)

Ta có:

\({n_{HCl}} = 2{n_{{H_2}}} = 2,7{\text{ mol}}\)

\( \to {m_{HCl}} = 2,7.36,5 = 98,55{\text{ gam}}\)

\( \to C{\% _{HCl}} = \frac{{98,55}}{{200}} = 49,275\% \)

BTKL:

\({m_{Al}} + {m_{dd\;{\text{HCl}}}} = {m_{dd{\text{ muối}}}} + {m_{{H_2}}}\)

\( \to 24,3 + 200 = {m_{dd{\text{ muối}}}} + 1,35.2\)

\( \to {m_{dd{\text{ muối}}}} = 221,6{\text{ gam}}\)

\({m_{AlC{l_3}}} = {n_{Al}} = 0,9{\text{ mol}}\)

\( \to {m_{AlC{l_3}}} = 0,9.(27 + 35,5.3) = 120,15{\text{ gam}}\)

\( \to C{\% _{AlC{l_3}}} = \frac{{120,15}}{{221,6}} = 54,67\% \)

Đáp án:

 

Giải thích các bước giải: