a) \(2SO_2+O_2\underrightarrow{t^o,V_2O_5}2SO_3\)
b)
\(n_{khí}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
=> \(n_{SO_2}=n_{O_2}=\dfrac{0,2}{2}=0,1\left(mol\right)\)
\(n_{khí.sau.pư}=0,2-0,2.10\%=0,18\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,1}{2}< \dfrac{0,1}{1}\) => Hiệu suất tính theo SO2
Gọi số mol SO2 pư là a (mol)
PTHH: \(2SO_2+O_2\underrightarrow{t^o}2SO_3\)
Bđ: 0,1 0,1
Pư: a--->0,5a--->a
Sau pư: (0,1-a) (0,1-0,5a) a
\(\Rightarrow\left(0,1-a\right)+\left(0,1-0,5a\right)+a=0,18\)
\(\Rightarrow a=0,04\left(mol\right)\)
\(\Rightarrow H=\dfrac{0,04}{0,1}.100\%=40\%\)
c)
hh sau pư chứa \(\left\{{}\begin{matrix}SO_2:0,06\left(mol\right)\\O_2:0,08\left(mol\right)\\SO_3:0,04\left(mol\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%V_{SO_2}=\dfrac{0,06}{0,18}.100\%=33,33\%\\\%V_{O_2}=\dfrac{0,08}{0,18}.100\%=44,44\%\\\%V_{SO_3}=100\%-33,33\%-44,44\%=22,23\%\end{matrix}\right.\)
