Gọi x, y là số mol Fe, Cu.
$\Rightarrow 56x+64y=8,8$ (1)
$n_{NO_2}=\dfrac{8,96}{22,4}=0,4(mol)$
$HNO_3$ dư nên $Fe\to Fe^{+3}+3e$
Bảo toàn e: $3n_{Fe}+2n_{Cu}=n_{NO_2}$
$\Rightarrow 3x+2y=0,4$ (2)
$(1)(2)\Rightarrow x=0,1; y=0,05$
$\%m_{Fe}=\dfrac{56x.100}{8,8}=63,6\%$
$\Rightarrow \%m_{Cu}=36,4\%$