Do $\text{iron}$ là $\text{Fe}$
$n_{Fe}=\dfrac{16,8}{56}=0,3(mol)$
Phương trình:
$3Fe+2O_2\xrightarrow{t^o}Fe_3O_4$
$\to n_{O_2}=0,2(mol)$
$n_{Fe_3O_4}=0,1(mol)$
Phương trình:
$2KClO_3\xrightarrow{t^o}2KCl+3O_2$
$a) \to V_{O_2}=0,2.22,4=4,48l$
$b) m_{Fe_3O_4}=0,1.232=23,2g$
$c)$ Ta có:
$n_{KClO_3}=\dfrac{2}{3}n_{O_2}$
$\to n_{KClO_3}=0,2.\dfrac{2}{3}$
$\to m_{KClO_3}=0,2.\dfrac{2}{3}.122,5=16,3g$