a, \(3Fe+2O_2\underrightarrow{t^0}Fe_3O_4\)
\(b,n_{Fe}=\dfrac{m}{M}=\dfrac{22,4}{56}=0,4\left(mol\right)\)
\(n_{O_2}=\dfrac{V}{22,4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
.................. \(3Fe+2O_2\underrightarrow{t^0}Fe_3O_4\)
Theo PTHH:\(3\left(mol\right)\)...\(2\left(mol\right)\)
Theo bài : \(0,4\left(mol\right)\)...\(0,4\left(mol\right)\)
Tỉ lệ : \(\dfrac{0,4}{3}< \dfrac{0,4}{2}\)
\(\Rightarrow\) O2 dư
Theo PTHH :
\(n_{O_2}=\dfrac{2}{3}.n_{Fe}=\dfrac{2}{3}.0,4=\dfrac{4}{15}\left(mol\right)\)
\(\Rightarrow V_{O_2}=n.22,4=\dfrac{4}{15}.22,4\sim5,97\left(l\right)\)
nFe = \(\dfrac{22,4}{56}\) =0,4 mol
nO2 = \(\dfrac{8,96}{22,4}\) = 0,4 mol
3Fe + 2O2 → Fe3O4 (1)
0,4 mol;0,4mol
theo pt(1) nFe hết ; nO2 dư
⇒nO2 = \(\dfrac{2}{3}\)nFe = 0,26 mol
⇒VO2 = 0,26 . 22,4 = 5,824 (l)
