\(n_S=\dfrac{16}{32}=0,5\left(mol\right)\)
\(n_{SO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PTHH: \(S+O_2\underrightarrow{t^o}SO_2\)
a. Theo PT ta có tỉ lệ: \(\dfrac{0,5}{1}>\dfrac{0,4}{1}\)
=> S dư. \(SO_2\) hết. => tính theo \(n_{SO_2}\)
b. Theo PT: \(n_{O_2}=n_{SO_2}=0,4\left(mol\right)\)
=> \(V_{O_2}=0,4.22,4=8,96\left(l\right)\)