\(n_{Fe}=\dfrac{5,6}{56}=0,1mol\\ 3Fe+8HNO_3\rightarrow3Fe\left(NO_3\right)_2+2NO+4H_2O\\ n_{Fe\left(NO_3\right)_2}=n_{Fe}=0,1mol\\ m_{Fe\left(NO_3\right)_2}=0,1.180=18g\\ n_{NO}=0,1\cdot\dfrac{2}{3}=\dfrac{1}{15}mol\\ V_{NO}=\dfrac{1}{15}\cdot22,4=\dfrac{112}{75}\approx1,49l\)