Đặt $2x$, $x$ (mol) là số mol Mg, Fe.
$\Rightarrow 24.2x+56x=10,4$
$\Leftrightarrow x=0,1$
$\Rightarrow n_{Mg}=0,2(mol); n_{Fe}=0,1(mol)$
$2Mg+O_2\buildrel{{t^o}}\over\to 2MgO$
$3Fe+2O_2\buildrel{{t^o}}\over\to Fe_3O_4$
$\Rightarrow n_{O_2}=\dfrac{1}{2}n_{Mg}+\dfrac{2}{3}n_{Fe}=\dfrac{1}{6}(mol)$
$\Rightarrow V_{O_2}=\dfrac{1}{6}.22,4=\dfrac{56}{15}l$
$\to V_{kk}=5V_{O_2}=18,67l$