\(n_{S_{ban.dau}}=0,15625\)
\(n_{O2_{ban.dau}}=\frac{6,4}{32}=0,2\left(mol\right)\)
\(S+O_2\rightarrow SO_2\)
\(\Rightarrow m=m_{SO2}=0,15625.64=10\left(g\right)\)
\(n_S=\frac{5}{32}=0,16\left(mol\right)\)
\(n_{O2}=\frac{6,4}{32}=0,2\left(mol\right)\)
\(\Rightarrow n_S< n_{O2}\Rightarrow\) O2 dư
\(PTHH:S+O_2\rightarrow SO_2\)
_______0,16______0,16___
\(\Rightarrow m=0,16.64=10,24\left(g\right)\)