Đáp án:
\(\begin{array}{l}
\% {m_{Al}} = 87,1\% \\
\% {m_{Mg}} = 12,9\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
8Al + 30HN{O_3} \to 8Al{(N{O_3})_3} + 3{N_2}O + 15{H_2}O\\
4Mg + 10HN{O_3} \to 4Mg{(N{O_3})_2} + {N_2}O + 5{H_2}O\\
b)\\
{n_{{N_2}O}} = \dfrac{{0,56}}{{22,4}} = 0,025\,mol\\
hh:Al(a\,mol),Mg(b\,mol)\\
\left\{ \begin{array}{l}
\dfrac{{3a}}{8} + \dfrac{b}{4} = 0,025\\
27a + 24b = 1,86
\end{array} \right.\\
\Rightarrow a = 0,06;b = 0,01\\
\% {m_{Al}} = \dfrac{{0,06 \times 27}}{{1,86}} \times 100\% = 87,1\% \\
\% {m_{Mg}} = 100 - 87,1 = 12,9\%
\end{array}\)