\(PTHH:S+O_2\underrightarrow{^{t^o}}SO_2\)
_______0,2___0,2____0,2
a)Ta có:
\(\left\{{}\begin{matrix}n_S=\frac{6,4}{32}=0,2\left(mol\right)\\n_{O2}=\frac{11,2}{16}=0,7\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\frac{n_{O2}}{1}=\frac{0,2}{1}< \frac{n_S}{1}=\frac{0,7}{1}\)
Nên O2 dư
Theo PTHH ta có nSO2=0,2(mol)
\(\Rightarrow V_{SO2}=0,2.22,4=4,48\left(l\right)\)
\(\Rightarrow n_{O2\left(pư\right)}=0,2\left(mol\right)\Rightarrow n_{O2\left(dư\right)}=0,7-0,2=0,5\left(mol\right)\)
\(\Rightarrow V_{O2}=0,5.22,4=11,2\left(l\right)\)
b)\(n_{O2\left(cd\right)}=0,2\left(mol\right)\Rightarrow V_{O2\left(cd\right)}=0,2.22,4=4,48\left(l\right)\)
\(\Rightarrow V_{kk}=\frac{4,48.100}{20}=22,4\left(l\right)\)
