$\overline{M}_{hh}=19.2=38$
Gọi x, y là số mol $NO$, $NO_2$
$\Rightarrow 30x+46y=38(x+y)$
$\Leftrightarrow x=y$
$\Rightarrow n_{NO}=n_{NO_2}=x$
$n_{Cu}=\dfrac{16}{64}=0,25(mol)$
Bảo toàn e: $2n_{Cu}=3n_{NO}+n_{NO_2}$
$\Rightarrow 2.0,25=3x+x$
$\Leftrightarrow x=0,125$
Bảo toàn N: $n_{HNO_3}=2n_{Cu(NO_3)_2}+n_{NO}+n_{NO_2}$
$=2n_{Cu}+x+x=0,75(mol)$
$\to C_{M_{HNO_3}}=\dfrac{0,75}{0,2}=3,75M$