$n_{Cu}=\dfrac{6,4}{64}=0,1(mol)$
$n_{H^+}=n_{HCl}=0,2.2=0,4(mol)$
$n_{NO_3^-}=n_{NaNO_3}=0,2(mol)$
$3Cu+8H^++2NO_3^-\to 3Cu^{2+}+2NO+4H_2O$
$\Rightarrow Cu$ hết
$n_{NO}=\dfrac{0,1.2}{3}=\dfrac{1}{15}(mol)$
$\Rightarrow V_{NO}=\dfrac{1}{15}.22,4=1,493l$
$n_{NO_3^-\text{dư}}=0,2-\dfrac{1}{15}=\dfrac{2}{15}(mol)<2n_{Cu^{2+}}$
Vậy sau phản ứng tạo thêm muối $CuCl_2$
$\Rightarrow n_{Cu(NO_3)_2}=0,5n_{NO3^-\text{dư}}=\dfrac{1}{15}(mol)$
$\to m_{Cu(NO_3)_2}=188.\dfrac{1}{15}=12,53g$
