Đáp án:

\(\% {m_{Al}} = 81,81\% ;\% {m_{Mg}} = 18,19\% \)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(4Al + 3{O_2}\xrightarrow{{}}2A{l_2}{O_3}\)

\(2Mg + {O_2}\xrightarrow{{}}2MgO\)

Ta có: \({n_{Mg}} = \frac{{2,4}}{{24}} = 0,1{\text{ mol}}\)

\(\to {n_{{O_2}}} = \frac{3}{4}{n_{Al}} + \frac{1}{2}{n_{Mg}} = \frac{{7,84}}{{22,4}} = 0,35{\text{ mol}} \to {{\text{n}}_{Al}} = 0,4{\text{ mol}} \to {{\text{m}}_{Al}} = 0,4.27 = 10,8{\text{ gam}}\)

\(\to \% {m_{Al}} = \frac{{10,8}}{{10,8 + 2,4}} = 81,81\%  \to \% {m_{Mg}} = 18,19\% \)

\(\begin{array}{l}
n_{Mg}=\frac{2,4}{24}=0,1(mol)\\
n_{O_2}=\frac{7,84}{22,4}=0,35(mol)\\
2Mg+O_2\xrightarrow{t^o}2MgO\\
4Al+3O_2\xrightarrow{t^o}2Al_2O_3\\
\to \frac{1}{2}n_{Mg}+\frac{3}{4}n_{Al}=n_{O_2}\\
\to 0,05+\frac{3}{4}n_{Al}=0,35\\
\to n_{Al}=0,4(mol)\\
\to m_{hh\,bđ}=0,4.27+2,4=13,2(g)\\
\to \%m_{Mg}=\frac{2,4}{13,2}.100\%\approx 18,18\%\\
\to \%m_{Al}=100-18,18=81,82\%
\end{array}\)