$m_{Fe}=5,6(g) \to n_{Fe}=\dfrac{5,6}{56}=0,1(mol)$
$\text{a) PTHH:}$
$3Fe + 2O_2 \xrightarrow{t^0} Fe_3O_4$
$\text{b) Ta có:}$
$n_{O_2}=0,1.\dfrac{2}{3}=\dfrac{1}{15}(mol)$
$\to V_{O_2}=\dfrac{1}{15}.22,4=1,49(l)$
$\text{c)}$
$n_{Fe_3O_4}=\dfrac{0,1.1}{3}=\dfrac{1}{30}(mol)$
$\to m_{Fe_3O_4}=232.\dfrac{1}{30}=7,73(g)$