Ta co pthh
3Fe + 2O2-to\(\rightarrow\) Fe3O4
Theo de bai ta co
nFe=\(\dfrac{33,6}{56}=0,6mol\)
Theo pthh
nO2=\(\dfrac{2}{3}nFe=\dfrac{2}{3}.0,6=0,4mol\)
\(\Rightarrow VO2_{\left(dktc\right)}\)=0,4.22,4=8,96 l
Theo pthh
nFe3O4=\(\dfrac{1}{3}nFe=\dfrac{1}{3}.0,6=0,2mol\)
\(\Rightarrow mFe3O4=0,2.232=46,4g\)