$n_{Cu}=\dfrac{6,4}{64}=0,1(mol)$
$n_{NO}+n_{NO_2}=\dfrac{2,24}{22,4}=0,1(mol)$
$n_{HNO_3}=0,35(mol)$
Nhận thấy: $2n_{Cu}+n_{NO}+n_{NO_2}=0,3<0,35$
Do đó $HNO_3$ dư, $Cu$ hết
$n_{Cu(NO_3)_2}=0,1(mol)$
$\to C_{M_{Cu(NO_3)_2}}=\dfrac{0,1}{0,35}=0,29M$
$n_{HNO_3\text{dư}}=0,35-0,3=0,05(mol)$
$\to C_{M_{HNO_3}}=\dfrac{0,05}{0,35}=0,14M$