Đáp án:
\(\% {m_{Al}} = 14,75\% \)
\(\% {m_{Mg}} = 85,25\% \)
Giải thích các bước giải:
\(\begin{array}{l}
Mg + \dfrac{1}{2}{O_2} \to MgO\\
4Al + 3{O_2} \to 2A{l_2}{O_3}\\
{n_{{O_2}}} = 0,4mol\\
{n_{Al}} = 0,1mol\\
\to {n_{{O_2}}} = \dfrac{1}{2}{n_{Mg}} + \dfrac{3}{4}{n_{Al}} = 0,4mol\\
\to {n_{Mg}} = 0,65mol\\
\to {m_{Mg}} = 15,6g\\
\to \% {m_{Mg}} = \dfrac{{15,6}}{{15,6 + 2,7}} \times 100\% = 85,25\% \\
\to \% {m_{Al}} = 14,75\%
\end{array}\)