\(Fe+4HNO_3\rightarrow Fe\left(NO_3\right)_3+NO+2H_2O\\ n_{NO}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)\\ Vì:\dfrac{0,5}{1}>\dfrac{0,1}{1}\Rightarrow Fedư\\ n_{HNO_3\left(p.ứ\right)}=4.n_{NO}=4.0,1=0,4\left(mol\right)\)