trước 9h30 nha các bạn cảm ơn nhìu nhìu

$a,\\ n_{NO}=\dfrac{3,36}{22,4}=0,15(mol)\\ Đặt\ \begin{cases}n_{Fe}=x(mol)\\n_{FeO}=y(mol)\end{cases}\\ \to 56x+72y=16,4\ (1)\\ Fe+4HNO_3\to Fe(NO_3)_3+2H_2O+NO\\ 3FeO+10HNO_3\to 3Fe(NO_3)_3+5H_2O+NO\\ n_{NO}=n_{Fe}+\dfrac{1}{3}.n_{FeO}=x+\dfrac{1}{3}y(mol)\\ \to x+\dfrac{1}{3}y=0,15\ (2)\\ \xrightarrow{\text{Từ (1),(2)}}\begin{cases}x=0,1\\y=0,15\end{cases}\\ \to \%m_{Fe}=\dfrac{0,1.56}{16,4}.100\%=34,15\%\\ \to \%m_{FeO}=\dfrac{0,15.72}{16,4}.100\%=65,85\%\\ b,\\ n_{HNO_3}=4.n_{Fe}+\dfrac{10}{3}.n_{FeO}\\ =0,9(mol)$