Đáp án:

\(\begin{array}{l}
a)\\
{d_{X/{H_2}}} = 20,4\\
b)\\
{n_{{N_2}}} = 0,06\,mol\\
{n_{{N_2}O}} = 0,24\,moll\\
c)\\
{N_{{N_2}}} = 1,2 \times {10^{22}}
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
hh\,X:{N_2}(a\,mol),{N_2}O(b\,mol)\\
{m_{{N_2}}}:{m_{{N_2}O}} = 7:44 \Leftrightarrow \dfrac{{28a}}{{44b}} = \dfrac{7}{{44}}\\
 \Leftrightarrow \dfrac{a}{b} = \dfrac{1}{4} \Rightarrow b = 4a\\
{m_{hh}} = 28a + 44b = 28a + 44 \times 4a = 204a(g)\\
{n_{hh}} = a + b = a + 4b = 5a\,mol\\
{M_{hh}} = \dfrac{{204a}}{{5a}} = 40,8\,g/mol\\
{d_{X/{H_2}}} = \dfrac{{40,8}}{2} = 20,4\\
b)\\
{n_{hh}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
{n_{{N_2}}} = \dfrac{{0,3}}{5} = 0,06\,mol\\
{n_{{N_2}O}} = 0,3 - 0,06 = 0,24\,moll\\
c)\\
{n_X} = \dfrac{{4,08}}{{40,8}} = 0,1\,mol\\
{n_{{N_2}}} = \dfrac{{0,1}}{5} = 0,02\,mol\\
{N_{{N_2}}} = 0,02 \times 6 \times {10^{23}} = 1,2 \times {10^{22}}\text{ phân tử}
\end{array}\)