Đáp án:
m dung dịch HCl=825 gam
Giải thích các bước giải:
a) \[\begin{gathered}
2Al + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2} \hfill \\
A{l_2}{O_3} + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}O \hfill \\
\end{gathered} \]
b) Ta có: \[{n_{{H_2}}} = \frac{{6,72}}{{22,4}} = 0,3{\text{ mol}}\]
Theo phản ứng:
\[{n_{{H_2}}} = \frac{3}{2}{n_{Al}} \to {n_{Al}} = 0,2mol \to {m_{Al}} = 0,2.27 = 5,4{\text{ gam}} \to {{\text{m}}_{A{l_2}{O_3}}} = 15,3{\text{ gam}} \to {{\text{n}}_{A{l_2}{O_3}}} = \frac{{15,3}}{{27.2 + 16.3}} = 0,15{\text{ mol}}\]
Ta có:
\[{n_{HCl}} = 3{n_{Al}} + 6{n_{A{l_2}{O_3}}} = 0,2.3 + 0,15.6 = 1,5{\text{ mol}} \to {{\text{n}}_{HCl{\text{ tham gia}}}} = 1,5.110\% = 1,65{\text{ mol}}\]
\[ \to {m_{HCl}} = 1,65.36,5 = 60,225{\text{ gam}} \to {{\text{m}}_{dd{\text{ HCl}}}} = \frac{{60,225}}{{7,3\% }} = 825{\text{ gam}}\]
