$a)$ Đặt $n_{Mg}=x(mol)\Rightarrow n_{Fe}=3x(mol)$

Ta có $m_{Mg}+m_{Fe}=12,6$

$\Rightarrow 24x+3x.56=12,6$

$\Rightarrow x=0,065625(mol)$

$PTHH:2Mg+O_2\xrightarrow{t^o}2MgO$

$3Fe+2O_2\xrightarrow{t^o}Fe_3O_4$

Theo PT: $n_{MgO}=n_{Mg}=0,065625(mol);n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=\dfrac{7}{320}(mol)$

$\Rightarrow m_{MgO}=0,65625.40=2,625(g);m_{Fe_3O_4}=\dfrac{7}{320}.232=5,075(g)$

$\Rightarrow m_A=2,625+5,075=7,7(g)$

$b)$ Theo PT: $n_{O_2}=\dfrac{1}{2}n_{Mg}+\dfrac{2}{3}n_{Fe}=\dfrac{49}{640}(mol)$

$\Rightarrow V_{O_2(p/ứ)}=\dfrac{49}{640}.22,4=1,715(lít)$

$\Rightarrow V_{O_2(dùng)}=1,715.(100\%+20\%)=2,058(lít)$ 

 `a)` 

`3Fe + 2O_2`$\xrightarrow{t°}$`Fe_3O_4` 

`2Mg + O_2`$\xrightarrow{t°}$`2MgO` 

`n_{Fe} = 3x(mol) → n_{Mg} = x(mol)`

`m =56 .3x + 24x = 12,6(g)`

`→ x = 0,065625(mol)`

`→ n_{Fe} = 0,196875(mol)` 

Theo pt: 

`n_{O_2} = 2/3n_{Fe} + 1/2n_{Mg}`

`= 2/3 .0,196875 + 1/2 . 0,065625`

`= 0,1640625(mol)`

Bảo toàn khối lượng: 

`m_r = 12,6 + 0,1640625 . 32 = 17,85(g)`

`b)` 

`V_{O_2} = {0,1640625.  22,4}/{120%} = 3,0625(l)`