$1,n_{Cu}=\dfrac{72.10^{23}}{6.10^{23}}=12(mol)\\ 2,\text{Đặt }n_{N}=a(mol);n_{NO_2}=b(mol)\\ \text{Ta có }M_X=\dfrac{30a+46b}{a+b}=20.M_{H_2}=20.2=40(g/mol)\\ \Rightarrow 30a+46b=40a+40b\\ \Rightarrow 10a=6b\Rightarrow a=\dfrac{3}{5}b=0,6b\\ \Rightarrow \%_{V_{NO}}=\dfrac{22,4a}{22,4(a+b)}.100\%=\dfrac{a}{a+b}.100\%=\dfrac{0,6b}{0,6b+b}.100\%=\dfrac{3}{8}.100\%=37,5\%$

Em tham khảo nha :

\(\begin{array}{l}
1)\\
{n_{Cu}} = \dfrac{{{N_{Cu}}}}{{{N_A}}} = \dfrac{{72 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 12mol\\
2)\\
{M_{hh}} = 20{M_{{H_2}}} = 40dvC\\
hh:NO(a\,mol),N{O_2}(b\,mol)\\
 \Rightarrow \dfrac{{30a + 46b}}{{a + b}} = 40\\
 \Rightarrow a = \dfrac{6}{{10}}b \Rightarrow a = 0,6b\\
\% {V_{NO}} = \dfrac{a}{{a + b}} \times 100\%  = a:(a + \dfrac{5}{3}a) \times 100\%  = 37,5\% 
\end{array}\)