$n_{H_2SO_4}=0,12(mol)$
$n_{H_2}=\dfrac{0,224}{22,4}=0,01(mol)$
$X(Fe, O)+H_2SO_4\to$ muối $+H_2+H_2O$
Bảo toàn $H$, ta có:
$2n_{H_2SO_4}=2n_{H_2}+2n_{H_2O}$
$\to n_{H_2O}=0,12-0,01=0,11(mol)$
Bảo toàn $O$ ta có:
$n_{O\text{trong X}}+4n_{H_2SO_4}=4n_{SO_4\text{trong muối}}+n_{H_2O}$
$\to n_{O\text{trong X}}=n_{H_2O}=0,11(mol)$
$\to m_{Fe\text{trong X}}=7,36-0,11.16=5,6g$
Vậy $m=5,6g$