Gọi x, y là số mol Fe, CuO.
$\Rightarrow 56x+80y=1,36$ (1)
Bảo toàn nguyên tố:
$n_{Fe(NO_3)_3}=n_{Fe}=x (mol)$
$n_{Cu(NO_3)_2}=n_{Cu}= y (mol)$
$\Rightarrow 242x+188y=4,3$ (2)
(1)(2)$\Rightarrow x=y=0,01$
Bảo toàn e: $3n_{Fe}=3n_{NO}$
$\Rightarrow n_{NO}=0,01(mol)$
$\to V=0,01.22,4=0,224l$
$\%m_{Fe}=\dfrac{0,01.56.100}{1,36}=41,18\%$
$\%m_{CuO}=58,82\%$
