$Fe_2O_3+3H_2\buildrel{{t^o}}\over\to 2Fe+3H_2O$
$Fe_2O_3+3CO\buildrel{{t^o}}\over\to 2Fe+3CO_2$
$\overline{M}_A=9,66.2=19,32$
$\Rightarrow \dfrac{n_{H_2}}{n_{CO}}=\dfrac{28-19,32}{19,32-2}=0,5=\dfrac{1}{2}$
Đặt $x$, $2x$ là số mol $H_2$, $CO$ trong A.
$n_{Fe}=\dfrac{16,8}{56}=0,3(mol)$
$\Rightarrow 1,5.0,3=x+2x$
$\Leftrightarrow x=0,15$
$\to V_{H_2}=0,15.22,4=3,36l; V_{CO}=0,15.2.22,4=6,72l$