Đáp án:
\(\begin{array}{l}
a)\\
\% Al = 52,94\% \\
\% Mg = 47,06\% \\
b)\\
{V_{{H_2}}} = 5,6l\\
c)\\
{V_{S{O_2}}} = 5,6l
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Al + 2{H_2}O + 2NaOH \to 2NaAl{O_2} + 3{H_2}\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,1mol\\
{m_{Al}} = n \times M = 0,1 \times 27 = 2,7g\\
{m_{Mg}} = 5,1 - 2,7 = 2,4g\\
\% Al = \dfrac{{2,7}}{{5,1}} \times 100\% = 52,94\% \\
\% Mg = 100 - 52,94 = 47,06\% \\
b)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
{n_{Mg}} = \dfrac{m}{M} = \dfrac{{2,4}}{{24}} = 0,1mol\\
{n_{{H_2}}} = \dfrac{3}{2}{n_{Al}} + {n_{Mg}} = 0,25mol\\
{V_{{H_2}}} = n \times 22,4 = 0,25 \times 22,4 = 5,6l\\
c)\\
2Al + 6{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
Mg + 2{H_2}S{O_4} \to MgS{O_4} + S{O_2} + 2{H_2}O\\
{n_{S{O_2}}} = \dfrac{3}{2}{n_{Al}} + {n_{Mg}} = 0,25mol\\
{V_{S{O_2}}} = n \times 22,4 = 0,25 \times 22,4 = 5,6l
\end{array}\)
