$n_{Fe}=16,8/56=0,3mol$
$PTHH :$
$3Fe+2O_2\overset{t^o}\to Fe_3O_4$
a/Theo pt :
$n_{O_2}=2/3.n_{Fe}=2/3.0,3=0,2mol$
$⇒V_{O_2}=0,2.22,4=4,48l$
b/Theo pt :
$n_{Fe_3O_4}=1/3.n_{Fe}=1/3.0,3=0,1mol$
$⇒m_{Fe_3O_4}=0,1.232=23,2g$
$c/V_{kk}=4,48.5=22,4l$
Chúc bạn học tốt😊😊