Ta có: \(n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\)
\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PT: \(S+O_2\underrightarrow{t^o}SO_2\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,3}{1}\), ta được O2 dư.
Theo PT: \(n_{SO_2}=n_S=0,1\left(mol\right)\)
\(\Rightarrow V_{SO_2}=0,1.22,4=2,24\left(l\right)\)