@Kaneki

$\rm {M_{H_2}=2}\hspace{3cm}{13}\\\hspace{1,4cm}\huge{\searrow}\hspace{2cm}\huge{\nearrow}\\\hspace{2,7cm}{M_A=15}\\\hspace{1,4cm}\huge{\nearrow}\hspace{2cm}\huge{\searrow}\\{M_{CO}=28}\hspace{3cm}{13}$

$\rm \dfrac{n_{H_2}}{n_{CO}}=\dfrac{13}{13}=\dfrac{1}{1}$ 

`=>`$\rm n_{H_2}=n_{CO}$ 

`->`$\rm V_{H_2}=V_{CO}$ 

Đặt $\rm x(l)=V_{H_2(thêm)}$ 

`->`$V_{H_2(sau)}=x+10(l)$

Ta có tỉ khối hơi giảm `2` lần `=>`$\rm M_A=7,5$ 

$\rm \dfrac{2.(x+10)+28.10}{x+20}=7,5$ 

`<=>`$\rm x=27,27(l)$ 

Suy ra lượng $\rm V_{H_2(thêm)}=27,27(l)$