Đáp án:
\( {m_{{H_2}}} = 0,3{\text{ gam}}\)
\( C{\% _{AlC{l_3}}} = 23,36\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2Al + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}\)
Ta có:
\({n_{Al}} = \frac{{2,7}}{{27}} = 0,1{\text{ mol}}\)
\( \to {n_{{H_2}}} = \frac{3}{2}{n_{Al}} = 0,15{\text{ mol}}\)
\( \to {m_{{H_2}}} = 0,15.2 = 0,3{\text{ gam}}\)
\({n_{HCl}} = 2{n_{{H_2}}} = 0,3{\text{ mol}}\)
\( \to {m_{HCl}} = 0,3.36,5 = 10,95{\text{ gam}}\)
\( \to {m_{dd{\text{ HCl}}}} = \frac{{10,95}}{{20\% }} = 54,75{\text{ gam}}\)
BTKL:
\({m_{Al}} + {m_{dd\;{\text{HCl}}}} = {m_{dd}} + {m_{{H_2}}}\)
\( \to 2,7 + 54,75 = {m_{dd}} + 0,3 \to {m_{dd}} = 57,15{\text{ gam}}\)
\( \to {n_{AlC{l_3}}} = {n_{Al}} = 0,1{\text{ mol}}\)
\( \to {m_{AlC{l_3}}} = 0,1.(27 + 35,5.3) = 13,35{\text{ gam}}\)
\( \to C{\% _{AlC{l_3}}} = \frac{{13,35}}{{57,15}} = 23,36\% \)
