Question

Answer 1

nFe = 126/56 = 2,25 (mol)

PTHH: 3Fe + 2O2 -> (t°) Fe3O4

Mol: 2,25 ---> 1,5 ---> 0,75

VO2 = 1,5 . 22,4 = 33,6 (l)

Answer 2

\(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{126}{56}=2,25mol\)

\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)

2,25    1,5                      ( mol )

\(V_{O_2}=n_{O_2}.22,4=1,5.22,4=33,6l\)