Question

Answer 1

a) \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

PTHH: 3Fe + 2O₂ --t^o--> Fe₃O₄

           0,2-->\(\dfrac{0,4}{3}\)------->\(\dfrac{0,2}{3}\)

=> \(V_{O_2}=\dfrac{0,4}{3}.22,4=\dfrac{224}{75}\left(l\right)\)

b) \(m_{Fe_3O_4}=\dfrac{0,2}{3}.232=\dfrac{232}{15}\left(g\right)\)