$PTPƯ:3H_2+Fe_2O_3\xrightarrow{t^o} 2Fe+3H_2O$
$n_{Fe}=\dfrac{22,4}{56}=0,4mol.$
$Theo$ $pt:$ $n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe}=0,2mol.$
$⇒m_{Fe_2O_3}=0,2.160=32g.$
$Theo$ $pt:$ $n_{H_2}=\dfrac{3}{2}n_{Fe}=0,6mol.$
$⇒V_{H_2}=0,6.22,4=13,44l.$
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