Đáp án:
\(\begin{array}{l}
{m_{Mg}} = 3,6g\\
{m_{MgO}} = 0,4g\\
{m_{{\rm{dd}}HN{O_3}}} = 59,85g\\
C{\% _{Mg{{(N{O_3})}_2}}} = 37,58\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
5Mg + 12HN{O_3} \to 5Mg{(N{O_3})_2} + {N_2} + 6{H_2}O\\
MgO + 2HN{O_3} \to Mg{(N{O_3})_2} + {H_2}O\\
{n_{{N_2}}} = \dfrac{{0,672}}{{22,4}} = 0,03mol\\
\Rightarrow {n_{Mg}} = 5{n_{{N_2}}} = 0,15mol\\
{m_{Mg}} = 0,15 \times 24 = 3,6g\\
{m_{MgO}} = 4 - 3,6 = 0,4g\\
{n_{MgO}} = \dfrac{{0,4}}{{40}} = 0,01mol\\
{n_{HN{O_3}}} = 12{n_{{N_2}}} + 2{n_{MgO}} = 0,38mol\\
{m_{HN{O_3}}} = 0,38 \times 63 = 23,94g\\
{m_{{\rm{dd}}HN{O_3}}} = \dfrac{{23,94 \times 100\% }}{{40\% }} = 59,85g\\
{m_{{N_2}}} = 0,03 \times 28 = 0,84g\\
{n_{Mg{{(N{O_3})}_2}}} = {n_{Mg}} + {n_{MgO}} = 0,15 + 0,01 = 0,16mol\\
{m_{Mg{{(N{O_3})}_2}}} = 0,16 \times 148 = 23,68g\\
{m_{{\rm{dd}}spu}} = 4 + 59,85 - 0,84 = 63,01g\\
C{\% _{Mg{{(N{O_3})}_2}}} = \dfrac{{23,68}}{{63,01}} \times 100\% = 37,58\%
\end{array}\)
