Sửa đề thành 89,6 l khí màu nâu, 896 thì 40 mol lận
Chất khí màu nâu đó là NO2
\(\Rightarrow n_{NO_2}=\dfrac{89,6}{22,4}=4\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{FeO}=b\left(mol\right)\end{matrix}\right.\)
PTHH:
Fe + 6HNO3 ---> Fe(NO3)3 + 3NO2 + 3H2O
a-------------------->a-------------->3a
FeO + 4HNO3 ---> Fe(NO3)3 + 2H2O + NO2
b------------------------>b------------------------>b
Hệ pt \(\left\{{}\begin{matrix}56a+72b=128\\3a+b=40\end{matrix}\right.\Leftrightarrow a=b=1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=1.56=56\left(g\right)\\m_{FeO}=1.72=72\left(g\right)\\m_{Fe\left(NO_3\right)_3}=\left(1+1\right).242=484\left(g\right)\end{matrix}\right.\)
