Đặt: \(O_2: a(mol); O_3: b(mol)\)
\(\frac{32a+48b}{a+b}=1,3.32 \Rightarrow 9,6a=6,4b \Rightarrow \frac{a}{b}=\frac{2}{3}\)
%\(m_{O_2}=\frac{2.32.100}{2.32+3.48}=30,77\)%
\(\Rightarrow\) %\(m_{O_3}=69,23\)%
Ta có:
\(32n_{O_2}+48.1,5n_{O_2}=20,8 \Rightarrow n_{O_2}=0,2 (mol) \Rightarrow n_{O_3}=0,3(mol)\)
Bảo toàn e: \(6n_{C_6H_6}.5=0,2.4+0,3.6 \Rightarrow n_{C_6H_6}=\frac{13}{150} \Rightarrow m_{C6H_6}=6,76(g)\)