\(n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\\
n_{O_2}=\dfrac{6,4}{32}=0,2\left(mol\right)\\
pthh:S+O_2\underrightarrow{t^o}SO_2\\
LTL:\dfrac{0,1}{1}< \dfrac{0,2}{1}\)
=> Oxi dư
\(n_{SO_2}=n_S=0,1\left(mol\right)\\
m_{SO_2}=0,1.64=6,4\left(g\right)\)
\(n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\)
\(n_{O_2}=\dfrac{6,4}{16}=0,4\left(mol\right)\)
\(S+O_2\underrightarrow{t^o}SO_2\)
\(\dfrac{n_S}{n_{O_2}}=\dfrac{0,1}{0,4}< 1\) vậy Oxi dư
\(S+O_2\underrightarrow{t^o}SO_2\)
\(\left(mol\right)0,1\rightarrow0,1\rightarrow0,1\)
\(m_{SO_2}=0,1.64=6,4\left(g\right)\)