Đáp án:
\(\% {m_{Al}} = 12\% ;\% {m_{Ag}} = 88\% \)
\({m_{dd{\text{ HCl}}}} = 24,09{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2Al + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}\)
Ta có:
\({n_{{H_2}}} = \frac{{1,344}}{{22,4}} = 0,06{\text{ mol}}\)
\( \to {n_{Al}} = \frac{2}{3}{n_{{H_2}}} = 0,04{\text{ mol}}\)
\( \to {m_{Al}} = 0,04.27 = 1,08{\text{ gam}}\)
\( \to \% {m_{Al}} = \frac{{1,08}}{9} = 12\% \to \% {m_{Ag}} = 88\% \)
\({n_{HCl}} = 2{n_{{H_2}}} = 0,12{\text{ mol}}\)
\( \to {n_{HCl{\text{ tham gia}}}} = 0,12.(100\% + 10\% ) = 0,132{\text{ }}mol\)
\( \to {m_{HCl}} = 0,132.36,5 = 4,818{\text{ gam}}\)
\( \to {m_{dd{\text{ HCl}}}} = \frac{{4,818}}{{20\% }} = 24,09{\text{ gam}}\)