$n_{FeO}=a(mol);n_{Fe}=b(mol)$

`->`$72a+56b=18.4$$(*)$

$n_{NO}+n_{NO_2}=\dfrac{11.2}{22.4}=0.5(mol)$ $(1)$

$m_{NO+NO_2}=21.4 \times 0.5=21.4(gam)$

`->`$30n_{NO}+46n_{NO_2}=21.4$ $(2)$

Từ $(1),(2)$ suy ra: $n_{NO}=0.1;n_{NO_2}=0.4$

Bảo toàn $e:$

$a+3b=0.1 \times 3+0.4=0.7$ $(**)$

Từ $(*),(**)$ suy ra: $a=0.1;b=0.2$

`=>`$\%m_{FeO}=\dfrac{0.1 \times 72}{18.4} \times 100\%=39.13\%$

$\%m_{Fe}=100\%-39.13\%=60.87\%$

`A:FeO:x(mol);Fe:y(mol)`

`72x+56y=18,4(1)`

`M_{khí}=21,4.2=42,8(g//mol)`

`->{n_{NO}}/{n_{NO_2}}={46-42,8}/{42,8-30}=1/4`

`n_{NO}+n_{NO_2}={11,2}/{22,4}=0,5(mol)`

`->n_{NO}=0,1(mol);n_{NO_2}=0,4(mol)`

Bte: `x+3y=3n_{NO}+n_{NO_2}=0,7(2)`

`(1)(2)->x=0,1;y=0,2`

`\%m_{FeO}={0,1.72}/{18,4}.100\%=39,13\%`

`\%m_{Fe}=100-39,13=60,87\%`

Bte: `n_{HNO_3}=4n_{NO}+2n_{NO_2}=1,2(mol)`