a,
$n_{\uparrow}=\dfrac{3,024}{22,4}=0,135(mol)$
Gọi $x$, $y$ là số mol $NO_2$, $O_2$
$\Rightarrow x+y=0,135$
$\overline{M}_{\uparrow}=21,6.2=43,2$
$\Rightarrow 46x+32y=0,135.43,2$
Giải hệ: $x=0,108; y=0,027$
$\%V_{NO_2}=\dfrac{0,108.100}{0,135}=80\%$
$\%V_{O_2}=20\%$
b,
Bảo toàn N: $n_{Cu(NO_3)_2\text{pứ}}=\dfrac{x}{2}=0,054(mol)$
$\to H=\dfrac{0,054.188.100}{11,28}=90\%$