Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{Mg}} = 22,22\% \\
\% {m_{Fe}} = 77,78\% \\
b)\\
{m_{{\rm{dd}}HN{O_3}}} = 27,3g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Mg + 4HN{O_3} \to Mg{(N{O_3})_2} + 2N{O_2} + 2{H_2}O\\
Fe + 6HN{O_3} \to Fe{(N{O_3})_3} + 3N{O_2} + 3{H_2}O\\
hh:Mg(a\,mol),Fe(b\,mol)\\
{n_{N{O_2}}} = \dfrac{{2,912}}{{22,4}} = 0,13\,mol\\
\left\{ \begin{array}{l}
24a + 56b = 2,16\\
2a + 3b = 0,13
\end{array} \right.\\
\Rightarrow a = 0,02;b = 0,03\\
\% {m_{Mg}} = \dfrac{{0,02 \times 24}}{{2,16}} \times 100\% = 22,22\% \\
\% {m_{Fe}} = 100 - 22,22 = 77,78\% \\
b)\\
{n_{HN{O_3}}} = 2{n_{N{O_2}}} = 0,26\,mol\\
{m_{{\rm{dd}}HN{O_3}}} = \dfrac{{0,26 \times 63}}{{60\% }} = 27,3g
\end{array}\)
