Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{Mg}} = 68,71\% \\
\% {m_{Al}} = 31,29\% \\
b)\\
{C_M}HN{O_3} = 0,8M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
3Mg + 8HN{O_3} \to 3Mg{(N{O_3})_2} + 2NO + 4{H_2}O\\
Al + 4HN{O_3} \to Al{(N{O_3})_3} + NO + 2{H_2}O\\
{n_{NO}} = \dfrac{{2,016}}{{22,4}} = 0,09\,mol\\
hh:Mg(a\,mol),Al(b\,mol)\\
24a + 27b = 2,934\\
\dfrac{{2a}}{3} + b = 0,09\\
\Rightarrow a = 0,084;b = 0,034\\
\% {m_{Mg}} = \dfrac{{0,084 \times 24}}{{2,934}} \times 100\% = 68,71\% \\
\% {m_{Al}} = 100 - 68,71 = 31,29\% \\
b)\\
{n_{HN{O_3}}} = 0,084 \times \dfrac{8}{3} + 0,034 \times 4 = 0,36\,mol\\
{C_M}HN{O_3} = \dfrac{{0,36}}{{0,45}} = 0,8M
\end{array}\)
