\(PTHH:S+O_2\rightarrow SO_2\)
\(n_S=\frac{3,2}{32}=0,1\left(mol\right)\)
\(n_{O2}=\frac{1,12}{22,4}=0,05\left(mol\right)\)
Vậy S dư , O2 hết
\(\rightarrow n_{SO2}=n_{O2}=0,05\left(mol\right)\)
\(\rightarrow V_{SO2}=0,05.22,4=1,12\left(l\right)\)