$n_{Cu}=x(mol);n_{Fe}=y(mol)$

$n_{NO}=0.15(mol)$

Bảo toàn $e:$

$2x+3y=3n_{NO}=0.15 \times 3=0.45(mol)$

Bảo toàn nguyên tố $Cu:$

$n_{Cu(NO_3)_2}=n_{Cu}=x(mol)$

Bảo toàn nguyên tố $Fe:$

$n_{Fe(NO_3)_3}=n_{Fe}=y(mol)$

`->`$188a+242b=39.5$ $(2)$

$(1),(2)$`=>`$x=0.12;y=0.07$

`=>`$a=64x+56y=11.6(gam)$

Đáp án:

$\rm n_{NO}=\dfrac{3,36}{22,4}=0,15\ (mol)$

Bán phản ứng:

$\rm 4H^++NO_3^-+3e\to NO+2H_2O$

Ta có:

$\rm n_{NO_3^-}=3n_{NO}=0,45\ (mol)$

$\rm a=m_Y-m_{NO_3^-}=39,5-0,45×62=11,6\ (g)$