$\rm A: O_2; O_3$

Giả sử trong $\rm A$ có $\rm a(mol)$ hỗn hợp khí

Gọi $\rm x, y$ lần lượt là số mol của $\rm O_2; O_3$

Ta có PT:

$\rm n_{O_2} + n_{O_3} = a$

$\rm x + y = a(1)$

Theo đề:

$\rm M_A = 2 . 18 = 36g/mol$

$\rm m_A = 36a(g)$

$\rm \Rightarrow 32x + 48y = 36a$

$\rm \Rightarrow \dfrac{8}{9}x + \dfrac{4}{3}y = a(2)$

Từ $\rm (1),(2)$

$\rm \Rightarrow x + y = \dfrac{8}{9}x + \dfrac{4}{3}y$

$\rm \Rightarrow x - \dfrac{8}{9}x = \dfrac{4}{3}y - y$

$\rm \Rightarrow \dfrac{9}{9}x - \dfrac{8}{9}x = \dfrac{4}{3}y - \dfrac{3}{3}y$

$\rm \Rightarrow \dfrac{1}{9}x = \dfrac{1}{3}y$

$\rm \Rightarrow \dfrac{ \dfrac{1}{9} }{ \dfrac{1}{9} }x = \dfrac{ \dfrac{1}{3} }{ \dfrac{1}{3} }y$

$\rm \Rightarrow x = 3y$

$\rm \Rightarrow n_A = a = y + 3y = 4y(mol)$

Thể tích mỗi khí trong hỗn hợp $\rm A$ là:

$\rm \text{%}V_{O_3} = \dfrac{y}{4y}. 100\text{%} = 25$`%`

$\rm \text{%}V_{O_2} = 100$`%`$\rm - 25$`%`$\rm = 75$`%`

áp dụng phương trình đường chéo 

$\rm {M_{O_2}}\hspace{5cm}{12}\\\hspace{0,5cm}\huge{\searrow}\hspace{2,5cm}\huge{\nearrow}\\\hspace{2,4cm}{36}\\\hspace{0,5cm}\huge{\nearrow}\hspace{2,5cm}\huge{\searrow}\\{M_{O_3}}\hspace{5cm}{4}$ 

suy ra $\dfrac{V_{O_2}}{V_{O_3}}=\dfrac{12}{3}=\dfrac{3}{1}$ 

`<=>`$V_{O_2}=3V_{O_3}$ 

$\%V_{O_3}=\dfrac{1}{4}.100\%=25\%$

`->`$\%V_{O_2}=100\%-25\%=75\%$