$a,PTPƯ:Fe+H_2SO_4\xrightarrow{} FeSO_4+H_2↑$
$n_{Fe}=\dfrac{11,2}{56}=0,2mol.$
$Theo$ $pt:$ $n_{H_2}=n_{Fe}=0,2mol.$
$⇒V_{H_2}=0,2.22,4=4,48l.$
$b,Theo$ $pt:$ $n_{H_2SO_4}=n_{Fe}=0,2mol.$
$⇒C\%_{H_2SO_4}=\dfrac{0,2.98}{200}.100\%=9,8\%$
$c,Theo$ $pt:$ $n_{FeSO_4}=n_{Fe}=0,2mol.$
$⇒C\%_{FeSO_4}=\dfrac{0,2.152}{11,2+200-(0,2.2)}.100\%=14,42\%$
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