a)\(\%S:\%O=40:60\)
=>\(n_S:n_O=\frac{40}{32}:\frac{60}{16}=1,25:3,75=1:3\)
CTHH:SO3
b)\(S+O2-->SO2\)
\(n_S=\frac{3,2}{32}=0,1\left(mol\right)\)
\(n_{O2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
= O2 dư
\(n_{SO2}=n_S=0,1\left(mol\right)\)
\(m_{SO2}=0,1.6,4=6,4\left(g\right)\)