$Fe_3O_4+4H_2\xrightarrow{{t^o}} 3Fe+4H_2O$
$CuO+H_2\xrightarrow{{t^o}} Cu+H_2O$
$FeO+H_2\xrightarrow{{t^o}} Fe+H_2O$
$Fe_2O_3+3H_2\xrightarrow{{t^o}} 2Fe+3H_2O$
$m_{H_2SO_4}=100.83,3\%=83,3g$
Sau khi hấp thụ $H_2O$, $m_{dd}=83,3:74,575\%=111,7g$
$\to m_{H_2O\rm thêm}=111,7-100=11,7g$
$\to n_{H_2O}=n_{H_2}=\dfrac{11,7}{18}=0,65(mol)$
BTKL:
$m=20,08+0,65.1-0,65.2=30,48g$