\(M_{hh}=2.M_{N_2}=2.28=56\left(\dfrac{g}{mol}\right)\\ Đặt:a=n_{O_2};b=n_{SO_2}\left(a,b>0\right)\\ M_{hh}=56\left(\dfrac{g}{mol}\right)\\ \Leftrightarrow\dfrac{32a+64b}{a+b}=56\\ \Leftrightarrow8b=24a\\ \Leftrightarrow\dfrac{a}{b}=\dfrac{8}{24}=\dfrac{1}{3}\\ \Rightarrow\%V_{O_2}=\dfrac{1}{1+3}.100\%=25\%\Rightarrow\%V_{SO_2}=75\%\)
pt đường chéo :
O2 :32 36
28
SO2:64 4
=>O2 :SO2 =9:1
->%O2=\(\dfrac{9}{10}.100=90\%\)
=>%SO2=10%