Question

Answer 1

\(n_{Fe}=\dfrac{16.8}{56}=0.3\left(mol\right)\)

\(3Fe+2O_2\underrightarrow{^{^{t^0}}}Fe_3O_4\)

\(0.3.......0.2\)

\(V_{O_2}=0.2\cdot22.4=4.48\left(l\right)\)

Answer 2

Ta có : \(\Rightarrow nFe=\dfrac{16,8}{56}=0,3\left(mol\right)\)

Theo phương trình hoá học :

\(nO_2=\dfrac{2}{3}n=\dfrac{2}{3}.0,3=0,2\left(mol\right)\)

\(V=O_2=0,2.22,4=4,48\left(l\right)\)