$2)$$n_{FeO}=n_{Fe_3O_4}=n_{Fe_2O_3}=x mol$
Áp dụng định luật bảo toàn electron ta có :
$2x+2n_{CO}=3n_{NO}\Rightarrow 2x+\frac{2.2,352}{22,4}=\frac{3.2,24}{22,4}\Rightarrow x=0,045 mol$
$\Rightarrow n_{HNO_3}=3n_{Fe(NO_3)_3}+n_{NO}=3(x+3x+2x)+0,1=0,91 mol$
$\Rightarrow m_{HNO_3}=57,33 gam$