Đáp án:
\( {V_{{H_2}}} = 6,72{\text{ lít}}\)
\({m_{dd{\text{ HCl}}}} = 150{\text{ gam}}\)
\(C{\% _{AlC{l_3}}} = 20,74\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2Al + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}\)
Ta có:
\({n_{Al}} = \frac{{5,4}}{{27}} = 0,2{\text{ mol}}\)
\( \to {n_{HCl}} = \frac{3}{2}{n_{Al}} = \frac{3}{2}.0,2 = 0,3{\text{ mol}}\)
\( \to {V_{{H_2}}} = 0,3.22,4 = 6,72{\text{ lít}}\)
\({n_{HCl}} = 2{n_{{H_2}}} = 0,6{\text{ mol}}\)
\( \to {m_{HCl}} = 0,6.36,5 = 21,9{\text{ gam}}\)
\({m_{dd{\text{ HCl}}}} = \frac{{21,9}}{{14,6\% }} = 150{\text{ gam}}\)
BTKL:
\({m_{Al}} + {m_{dd{\text{ HCl}}}} = {m_{dd{\text{ muối}}}} + {m_{{H_2}}}\)
\( \to 5,4 + 150 = {m_{dd}} + 0,3.2 \to {m_{dd}} = 154,8{\text{ gam}}\)
\({n_{AlC{l_3}}} = {n_{Al}} = 0,2{\text{ mol}}\)
\( \to {m_{AlC{l_3}}} = 0,2.(27.2 + 35,5.3) = 32,1{\text{ gam}}\)
\( \to C{\% _{AlC{l_3}}} = \frac{{32,1}}{{154,8}} = 20,74\% \)