\(n_{Fe_3O_4}=\frac{m_{Fe_3O_4}}{M_{Fe_3O_4}}=\frac{46,4}{352}=0,2mol\)
`a,` \(PTHH:3Fe+2O_2\underrightarrow{t^0}Fe_3O_4\)
\(Theo_{pthh}:3--2---1\left(mol\right)\)
\(Theo_{Đề}:x--y---0,2\left(mol\right)\)
\(\Rightarrow n_{Fe}=x=\frac{0,2\cdot3}{1}=0,6mol\)
\(\Rightarrow m_{Fe}=n_{Fe}\cdot M_{Fe}=0,6\cdot56=33,6g\)
\(b,n_{O_2}=y=\frac{0,2\cdot2}{1}=0,4mol\)
\(\Rightarrow V_{O_2}=n_{O_2}\cdot22,4=0,4\cdot22,4=8,96l\)