a) PTHH: $Fe+H_2SO_4→FeSO_4+H_2↑$
$n_{H_2}=\dfrac{4,48}{22,4}=0,2(mol)$
$n_{Fe}=n_{H_2}=0,2(mol)$
⇒ $m=m_{Fe}=0,2×56=11,2(g)$
b) $2Fe+6H_2SO_{4(đ)}\buildrel{{t^o}}\over\longrightarrow Fe_2(SO_4)_3+3SO_2↑+6H_2O$
Ta có: $n_{H_2SO_4}=3n_{Fe}=3×0,2=0,6(mol)$
→ $m_{H_2SO_4}=0,6×98=58,8(g)$
$n_{SO_2}=\dfrac{3}{2}n_{Fe}=\dfrac{3}{2}×0,2=0,3(mol)$
→ $m_{SO_2}=0,3×64=19,2(g)$
$n_{Fe_2(SO_4)_3}=\dfrac{1}{2}n_{Fe}=\dfrac{1}{2}×0,2=0,1(mol)$
→ $m_{Fe_2(SO_4)_3}=0,1×400=40(g)$
Mặt khác ta lại có: $m_{ddH_2SO_4}=\dfrac{58,8×100}{98}=60(g)$
$m_{\text{dd sau p/ứ}}=11,2+60-19,2=52(g)$
⇒ $C$%$_{Fe_2(SO_4)_3}=\dfrac{40}{52}×100$% $=$ $76,92$%