Đáp án:
\(\% {m_{Al}} = 57,45\% ; \% {m_{Mg}} = 42,55\% \)
\({m_{muối}} = 6,38gam\)
\( {V_{HCl}} = 0,07{\text{ lít}}\)
\( {m_{dd{\text{ HCl}}}} = 13,81{\text{ gam}}\)
Giải thích các bước giải:
Gọi số mol \(Al;Mg\) lần lượt là \(x;y\)
\( \to 27x + 24y = 1,41\)
Cho hỗn hợp tác dụng với \(HCl\)
\(2Al + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}\)
\(Mg + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2}\)
Ta có:
\({n_{{H_2}}} = \frac{{1,568}}{{22,4}} = 0,07{\text{ mol = }}\frac{3}{2}{n_{Al}} + {n_{Mg}} = 1,5x + y\)
GIải được: \(x=0,03;y=0,025\)
\( \to {m_{Al}} = 0,03.27 = 0,81{\text{ gam;}}{{\text{m}}_{Mg}} = 0,025.24 = 0,6{\text{ gam}}\)
\( \to \% {m_{Al}} = \frac{{0,81}}{{1,41}} = 57,45\% \to \% {m_{Mg}} = 42,55\% \)
\({n_{AlC{l_3}}} = {n_{Al}} = 0,03{\text{ mol;}}{{\text{n}}_{MgC{l_2}}} = {n_{Mg}} = 0,025{\text{ mol}}\)
\( \to {m_{muối}} = {m_{AlC{l_3}}} + {m_{MgC{l_2}}}\)
\( = 0,03.(27 + 35,5.3) + 0,025.(24 + 35,5.2) = 6,38{\text{ gam}}\)
\({n_{HCl}} = 2{n_{{H_2}}} = 0,07.2 = 0,14{\text{ mol}}\)
\( \to {V_{HCl}} = \frac{{0,14}}{2} = 0,07{\text{ lít}}\)
\({m_{HCl}} = 0,14.36,5 = 5,11{\text{ gam}}\)
\( \to {m_{dd{\text{ HCl}}}} = \frac{{5,11}}{{37\% }} = 13,81{\text{ gam}}\)
