\(4R+nO_2\rightarrow2R_2O_n\)
\(n_{O_2}=\frac{0,7}{22,4}=0,03125\left(mol\right)\)
\(TheoPT:n_R=\frac{4}{n}n_{O_2}=\frac{0,125}{n}\left(mol\right)\)
\(M_R=\frac{1}{\frac{0,125}{n}}=8n\)
Ta thấy : \(n=4\Rightarrow M_R=32\left(S\right)\)
\(\Rightarrow CT:S_2O_{\text{4}}=SO_2\)