\(n_S=\dfrac{6,4}{32}=0,2\left(mol\right)\)
\(n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
PTHH: S + O2 --to--> SO2
Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,5}{1}\) => S hết, O2 dư
PTHH: S + O2 --to--> SO2
0,2-->0,2------->0,2
=> \(m_{SO_2}=0,2.64=12,8\left(g\right)\)
\(m_{O_2\left(dư\right)}=\left(0,5-0,2\right).32=9,6\left(g\right)\)
