Đáp án+Giải thích các bước giải:

$4Al+3O_{2}\xrightarrow[]{t^{o}}2Al_{2}O_{3}(1)$

$3Fe+2O_{2}\xrightarrow[]{t^{o}}Fe_{3}O_{4}(2)$

`n_{Al}=2.n_{Al_{2}O_{3}}=2.(10,2)/(102)=0,2(mol)`

`→m_{Al}=0,2.27=5,4(g)`

`→m_{Fe}=13,8-5,4=8,4(g)`

`→n_{Fe}=(8,4)/(56)=0,15(mol)`

`PT_{1}→n_{O_{2}(1)}=3/4.n_{Al}=(3)/(4).0,2=0,15(mol)`

`PT_{2}→n_{O_{2}(2)}=2/3.n_{Fe}=(2)/(3).0,15=0,1(mol)`

`→n_{O_{2}(1)(2)}=0,15+0,1=0,25(mol)`

`→V_{O_{2}}=0,25.24,79=6,1975(l)`

`4Al+3O_2` $\xrightarrow{t^o}$ `2Al_2O_3`

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

`n_{Al_2O_3}={10,2}/{102}=0,1(mol)`

Theo pt:

`n_{Al}=2n_{Al_2O_3}=0,2(mol)`

`n_{Fe}={13,8-0,2.27}/{56}=0,15(mol)`

Theo pt:

`n_{O_2}=3/4n_{Al}+2/3n_{Fe}=0,25(mol)`

`V_{O_2}=0,25.24,79=6,1975(L)`