Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{Fe}} = 60,87\% \\
\% {m_{Mg}} = 39,13\% \\
b){m_{muối}} = 9,28g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
Fe + 4HN{O_3} \to Fe{(N{O_3})_3} + NO + 2{H_2}O\\
3Mg + 8HN{O_3} \to 3Mg{(N{O_3})_2} + 2NO + 4{H_2}O\\
\left\{ \begin{array}{l}
56a + 24b = 1,84\\
a + \dfrac{2}{3}b = 0,04
\end{array} \right. \to \left\{ \begin{array}{l}
a = 0,02\\
b = 0,03
\end{array} \right.\\
\to {n_{Fe}} = 0,02mol\\
\to {n_{Mg}} = 0,03mol\\
a)\\
\% {m_{Fe}} = \dfrac{{0,02 \times 56}}{{1,84}} \times 100\% = 60,87\% \\
\% {m_{Mg}} = 100\% - 60,87\% = 39,13\% \\
b)\\
{n_{Fe{{(N{O_3})}_3}}} = {n_{Fe}} = 0,02mol \to {m_{Fe{{(N{O_3})}_3}}} = 4,84g\\
{n_{Mg{{(N{O_3})}_2}}} = {n_{Mg}} = 0,03mol \to {m_{Mg{{(N{O_3})}_2}}} = 4,44g\\
\to {m_{muối}} = 9,28g
\end{array}\)
