Đáp án + Giải thích các bước giải:

$a\big)$

$n_{H_2}=\frac{5,6}{22,4}=0,25(mol)$

$2Al+6HCl\to 2AlCl_3+3H_2$

$Mg+2HCl\to MgCl_2+H_2$

Theo PT: $n_{HCl}=2n_{H_2}=0,5(mol)$

$\to m_{HCl}=0,5.36,5=18,25(g)$

$b\big)$

Đặt $n_{Al}=x(mol);n_{Mg}=y(mol)$

$\to 27x+24y=5,1(1)$

Theo PT: $1,5x+y=0,25(2)$

Từ $(1)(2)\to x=y=0,1(mol)$

$\to \%m_{Al}=\frac{0,1.27}{5,1}.100\%=52,94\%$

$\to \%m_{Mg}=100\%-52,94\%=47,06\%$

$c\big)$

Theo PT: $n_{MgCl_2}=y=0,1(mol);n_{AlCl_3}=x=0,1(mol)$

$\to \%m_{MgCl_2}=\frac{0,1.95}{0,1.95+0,1.133,5}.100\%\approx 41,58\%$

$\to \%m_{AlCl_3}=100\%-41,58\%=58,42\%$

Đáp án:

\({{\text{m}}_{HCl}} = 18,25{\text{ gam}}\)

\( {\text{% }}{{\text{m}}_{Mg}} = 47\% ; \% {m_{Al}} = 53\% \)

\( \% {m_{MgC{l_2}}} = 41,6\% ;\% {m_{AlC{l_3}}} = 58,4\% \)

Giải thích các bước giải:

 Gọi số mol \(Mg;Al\) lần lượt là \(x;y\)

\( \to 24x+27y=5,1\)

Phản ứng xảy ra:

\(Mg + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2}\)

\(2{\text{A}}l + 6HCl\xrightarrow{{}}2{\text{A}}lC{l_3} + 3{H_2}\)

Ta có:

\({n_{{H_2}}} = {n_{Mg}} + \frac{3}{2}{n_{Al}} = x + 1,5y = \frac{{5,6}}{{22,4}} = 0,25{\text{ mol}}\)

Giải được: \(x=y=0,1\)

\({n_{HCl}} = 2{n_{{H_2}}} = 0,5{\text{ mol}} \to {{\text{m}}_{HCl}} = 0,5.36,5 = 18,25{\text{ gam}}\)

\({m_{Mg}} = 0,1.24 = 2,4{\text{ gam}} \to {\text{% }}{{\text{m}}_{Mg}} = \frac{{2,4}}{{5,1}} = 47\%  \to \% {m_{Al}} = 53\% \)

\({n_{MgC{l_2}}} = {n_{Mg}} = 0,1;{n_{AlC{l_3}}} = {n_{Al}} = 0,1\)

\( \to {m_{MgC{l_2}}} = 0,1.95 = 9,5{\text{ gam;}}{{\text{m}}_{AlC{l_3}}} = 0,1.133,5 = 13,35{\text{ gam}}\)

\( \to \% {m_{MgC{l_2}}} = \frac{{9,5}}{{9,5 + 13,35}} = 41,6\%  \to \% {m_{AlC{l_3}}} = 58,4\% \)