Đáp án  + Giải thích các bước giải:

$n_{Fe} = \dfrac{m}{M} = \dfrac{11,2}{56} = 0,2\text{ mol}$

$n_{H_2} = \dfrac{V}{22,4} = \dfrac{2,24}{22,4} = 0,1\text{ mol}$

 PTHH $(I)$:

$Fe_2O_3 + 3CO \xrightarrow{t^o} 2Fe + 3CO_2$

PTHH$(II)$:

$Fe + 2HCl \longrightarrow FeCl_2 + H_2$

$Fe_2O_3 +6HCl \longrightarrow 2FeCl_3 + 3H_2O$

Theo PTHH$(II)$:
$n_{Fe(hh)} = n_{H_2} = 0,1\text{ mol}$
$\Rightarrow m_{Fe(hh)} = n . M = 0,1 . 56 = 5,6g$

$\Rightarrow m_{Fe(\text{sp pứ CO})} = 11,2 - 5,6 = 5,6g$

$\Rightarrow n_{Fe(\text{sp pứ CO})} = \dfrac{m}{M} = \dfrac{5,6}{56} = 0,1\text{ mol}$

Theo PTHH$(I)$:
$n_{Fe(\text{sp pứ CO})} = 2n_{Fe_2O_3} = 0,1\text{ mol}$

$\Rightarrow n_{Fe_2O_3} = \dfrac{1}{2} . 0,1 = 0,05\text{ mol}$

$\Rightarrow m_{Fe_2O_3} = n . M = 0,05 . 160 = 8g$

$\Rightarrow m_{hh} = 8 + 5,6 = 13,6g$

$\Rightarrow \%_{Fe(hh)} = \dfrac{5,6}{13,6} . 100\% \approx 41,18\%$
$\%_{Fe_2O_3} = 100 - 41,18 = 58,82\%$