$m_{\text{tăng}}=4,8g=m_O$
$\to n_O=\dfrac{4,8}{16}=0,3(mol)$
$n_{SO_2}=\dfrac{8,4}{22,4}=0,375(mol)$
Gọi chung hh $A$ là $Fe_xO_y$
$\to n_A=\dfrac{0,3}{y}(mol)$
$2Fe_xO_y+(6x-2y)H_2SO_4\to xFe_2(SO_4)_3+(3x-2y)SO_2+(6x-2y)H_2O$
$\to \dfrac{0,3}{y}.\dfrac{3x-2y}{2}=0,375$
$\to 0,3(3x-2y)=0,375.2y$
$\to 0,9x=1,35y$
$\to n_{Fe}: n_O=x:y=3:2$
$\to n_{Fe}=1,5n_O=0,45(mol)$
Vậy $m=0,45.56=25,2g$