a)
$2Fe+6H2SO4--->Fe2(SO4)3+3SO2+6H2O$
$nSO2=6,72/22,4=0,3(mol)$
$nFe=2/3nSO2=0,2(mol)$
=>$mFe=0,2.56=11,2(g)$
=>$mFeO=16,4-11,2=4,2(g)$
b)
$m dd sau pư=m Fe+mdd H2SO4-mSO2=11,2+200-0,3.64=192(g)$
$nFe2(SO4)3=1/3nSO2=0,1(mol)$
=>$mFe2(SO4)3=0,1.400=40(g)$
=>C%Fe2(SO4)3=40/192.100%=20,8%