$a)\ 3Fe+2O_2 \xrightarrow{t^o}Fe_3O_4$
$b)$
$\text{Ta có:}\ n_{Fe} = \dfrac{m}{M} = \dfrac{126}{56} = 2,25\ mol$
$⇒ n_{O_2} = \dfrac{2}{3}n_{Fe} = \dfrac{2}{3}.2,25 = 1,5\ mol⇒V_{O_2} =33,6\ l$
$c)$
$\text{PTHH:} 2KClO_3\xrightarrow{t^o}2KCl+3O_2$
$\text{Theo PTHH:}\ n_{KClO_3}=\dfrac{2}{3}n_{O_2} = \dfrac{2}{3}.1,5 = 1\ mol$
$⇒ m_{KClO_3}=n.M = 1. 122,5 = 122,5\ g$