$m_{Fe}=0,3m>(m-0,75m)\Rightarrow Fe,Cu du$$\Rightarrow dd X:Fe(NO_3)_2$
$3Fe+8HNO_3\rightarrow 3Fe(NO_3)_2+2NO\uparrow+4H_2O$
$x \frac{8x}{3} \frac{2x}{3}$
$Fe+4HNO_3\rightarrow Fe(NO_3)_2+2NO_2\uparrow+2H_2O$
$y 4y 2y$
$\Rightarrow \begin{cases}\frac{8x}{3}+4y=\frac{44,1}{63} \\\frac{2x}{3}+2y=\frac{5,6}{22,4} \end{cases}\Rightarrow \begin{cases}x=0,15 mol \\ y=0,075 mol \end{cases}\Rightarrow 0,25m=56(x+y)$
$\Rightarrow m=50,4 gam$