$a,PTPƯ:Fe_3O_4+4H_2\xrightarrow{t^o} 3Fe+4H_2O$
$n_{Fe}=\dfrac{33,6}{56}=0,6mol.$
$Theo$ $pt:$ $n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=0,2mol.$
$⇒m_{Fe_3O_4}=0,2.232=46,4g.$
$b,Theo$ $pt:$ $n_{H_2}=\dfrac{4}{3}n_{Fe}=0,8mol.$
$⇒V_{H_2}=0,8.22,4=17,92l.$
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