a,b,

Oxit bazo :

`FeO:` sắt (II) oxit

`BaO :` bari oxit

Oxit axit :

`P_2O_5 : ` diphotpho pentaoxit

`SO_3` : lưu huỳnh trioxit

________________________

(bazo):

`Fe(OH)_3` : sắt (III) hidroxit

Câu 2 :

`n_{Al}=(5,4)/(27)=0,2(mol)`

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

a,Từ PTHH :`n_{Al_2O_3}= 1/2 n_{Al} = 1/2 .0,2=0,1(mol)`

`⇒m_{Al_2O_3}=0,1.102=10,2(g)`

b,Từ PTHH : `n_{O_2}= 3/4 n_{Al} = 3/4 . 0,2=0,15(mol)`

`⇒V_{O_2}=0,15.22,4=3,36(lít)`

$\text{Đáp án+Giải thích các bước giải:}$

$\text{Bài 1:}$

$\text{A và B}$

$\text{Oxi axit:}$

$\text{$P_{2}O_{5}$: Điphotpho pentaoxit}$

$\text{$SO_{3}$: Lưu huỳnh trioxit}$

$\text{Oxit bazơ:}$

$\text{FeO: Sắt (II) oxit}$

$\text{BaO: Bari oxit}$

$\text{Bazơ:}$

$\text{$Fe(OH)_{3}$: Sắt (III) Hydroxit}$

$\text{Bài 2:}$

$\text{4Al + $3O_{2}$$\xrightarrow{t^o}$$2Al_{2}O_{3}$}$

$\text{A.}$

$\text{Ta có:}$

$\text{$n_{Al}$=$\dfrac{m_{Al}}{M_{Al}}$=$\dfrac{5,4}{27}$= 0,2 (mol)}$

$\text{Theo PT ta có: $n_{Al_{2}O_{3}}$=$\dfrac{2}{4}$$n_{Al}$=$\dfrac{2}{4}$ × 0,2= 0,1 (mol)}$

$\text{⇒ $m_{Al_{2}O_{3}}$=$n_{Al_{2}O_{3}}$ × $M_{Al_{2}O_{3}}$=0,1 × 102= 10,2 (g)}$

$\text{B.}$

$\text{Theo PT ta có: $n_{O_{2}}$=$\dfrac{3}{4}$$n_{Al}$=$\dfrac{3}{4}$ × 0,2= 0,15 (mol)}$

$\text{⇒ $V_{O_{2}}$=$n_{O_{2}}$ × 22,4=0,15 × 22,4= 3,36 (lít)}$