Đáp án + Giải thích các bước giải:

 $n_{SO_2} = \dfrac{4,48}{22,4} = 0,2\text{ mol}$

$n_{H_2} = \dfrac{2,24}{22,4} = 0,1\text{ mol}$

PTHH:
$2Al + 6H_2SO_{4(đ,n)} \longrightarrow Al_2(SO_4)_3 + 3SO_2 \uparrow + 6H_2O(1)$

$2FeO + 4H_2SO_{4(đ,n)} \longrightarrow Fe_2(SO_4)_3 + SO_2\uparrow  + 4H_2O(2)$

$2Al + 3H_2SO_4 \longrightarrow Al_2(SO_4)_3 + 3H_2(3)$

$FeO + H_2SO_4 \longrightarrow FeSO_4 + H_2O(4)$

Theo PTHH(3):

$n_{Al} = \dfrac{2}{3}n_{H_2} = \dfrac{2}{3} . 0,1 = \dfrac{1}{15}\text{ mol}$

$\Rightarrow m_{Al} = \dfrac{1}{15} . 27 = 1,8g$

Theo PTHH(1, 2):

$n_{SO_2} = \dfrac{3}{2}n_{Al} + \dfrac{1}{2}n_{FeO} = \dfrac{3}{2} . \dfrac{1}{15} + \dfrac{1}{2}n_{FeO} = 0,1 + \dfrac{1}{2}n_{FeO} = 0,2$

$\Rightarrow n_{FeO} = (0,2 - 0,1) . 2 = 0,2\text{ mol}$

$\Rightarrow m_{FeO} = 0,2 . 72 = 14,4g$

$\Rightarrow \%_{FeO} = \dfrac{14,4}{14,4 + 1,8} . 100\% \approx 88,89\%$ 

`n_(SO_2)=(4,48)/(22,4)=0,2(mol)`

`n_(H_2)=(2,24)/(22,4)=0,1(mol)`

`2Al+3H_2SO_(4_l)->Al_2(SO_4)_3+3H_2`

`n_(Al)=2/3n_(H_2)=2/3 .0,1=1/15(mol)`

Quá trình oxi hóa:

$Al→\mathop{Al}\limits^{+3}+3e$

`1/15->1/15->0,2(mol)`

$\mathop{Fe}\limits^{+2}→\mathop{Fe}\limits^{+3}+e$

`x->x->x(mol)`

Quá trình khử:

$\mathop{S}\limits^{+6}+2e→\mathop{S}\limits^{+4}$

`0,2->0,4->0,2(mol)`

`BT(e): 0,2+x=0,4=>x=0,2(mol)`

`=>n_(FeO)=0,2(mol)`

`=>m_(FeO)=0,2.72=14,4(g)`

`m_(hh)=1/15 .27+14,4=16,2(g)`

`=>C%_(FeO)=(14,4)/(16,2).100%=88,89%`