\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
PT: 3Fe + 2O2 ➝ Fe3O4
mol 0,15 ➝ 0,05
\(m_{Fe_3O_4}=0,05.232=11,6\left(g\right)\)
Số sắt nguyên chất là: 168.(100% - 20%) = 134,4(g)
\(n_{Fe}=\dfrac{134,4}{56}=2,4\left(mol\right)\)
PT: 3Fe + 2O2 ➝ Fe3O4
mol 2,4 ➝ 1,6
\(V_{O_2\left(đktc\right)}=1,6.22,4=35,84\left(l\right)\)