\(1.n_S=\dfrac{6,4}{32}=0,2mol;n_{O_2}=\dfrac{3,36}{22,4}=0,15mol\\ PTHH:S+O_2\underrightarrow{t^0}SO_2\)
Ta có: \(\dfrac{n_S}{1}=\dfrac{0,2}{1}>\dfrac{n_{O_2}}{1}=\dfrac{0,15}{1}\) => oxi phản ứng hết, lưu huỳnh dư.
\(m_{S\left(dư\right)}=32\left(0,2-0,15\right)=1,6\left(g\right)\)
\(n_{SO_2}=n_{O_2}=0,15mol\\ \Rightarrow V=0,15.22,4=3,36\left(L\right)\)
\(2.n_S=\dfrac{6,4}{32}=0,2mol;n_{SO_2}=\dfrac{9,6}{22,4}=0,43mol\\ PTHH:S+O_2\underrightarrow{t^0}SO_2\\ Ta.có:\dfrac{n_S}{1}=\dfrac{0,2}{1}< \dfrac{n_{SO_2}}{1}=\dfrac{0,43}{1}\Rightarrow S.cháy.hết.\\ n_{O_2}=n_S=0,2\Rightarrow V_{O_2}=0,2.22,4=4,48\left(L\right)\)
