Đáp án:
\(\begin{array}{l}
\% Al = 42,86\% \\
\% Mg = 57,14\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
4Al + 3{O_2} \to 2A{l_2}{O_3}(1)\\
2Mg + {O_2} \to 2MgO(2)\\
{n_{{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{n_{Al}} = \dfrac{m}{M} = \dfrac{{2,7}}{{27}} = 0,1mol\\
{n_{{O_2}(1)}} = \dfrac{3}{4}{n_{Al}} = 0,075mol\\
{n_{{O_2}(2)}} = {n_{{O_2}}} - {n_{{O_2}(1)}} = 0,15 - 0,075 = 0,075mol\\
{n_{Mg}} = 2{n_{{O_2}(2)}} = 0,15mol\\
{m_{Mg}} = n \times M = 0,15 \times 24 = 3,6g\\
\% Al = \dfrac{{2,7}}{{2,7 + 3,6}} \times 100\% = 42,86\% \\
\% Mg = 100 - 42,86 = 57,14\%
\end{array}\)
