\(n_{Fe}=\dfrac{10,08}{56}=0,18\left(mol\right)\\ n_{NO}=\dfrac{3,024}{22,4}=0,135\left(mol\right)\\ Fe+4HNO_3\rightarrow Fe\left(NO_3\right)_3+NO+2H_2O\\ Fe_{dư}+2Fe\left(NO_3\right)_3\rightarrow3Fe\left(NO_3\right)_2\\ n_{Fe\left(dư\right)}=0,18-0,135=0,045\left(mol\right)\\ n_{Fe\left(NO_3\right)_2}=3.0,045=0,135\left(mol\right)\\ \Rightarrow m_{Fe\left(NO_3\right)_2}=180.0,135=24,3\left(g\right)\\ \Rightarrow Chọn.B\)
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