a, \(n_X=0,6\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{SO2}:a\left(mol\right)\\n_{O2}:b\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=0,6\\\frac{64a+32b}{0,6}=24.2=48\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,3\\b=0,3\end{matrix}\right.\)
\(\Rightarrow\%_{SO2}=\%_{O2}=50\%\)
\(2SO_2+O_2⇌2SO_3\)
\(n_Y=0,5\left(mol\right)\)
\(\left\{{}\begin{matrix}n_{SO2\left(pư\right)}=2a\\n_{O2\left(pư\right)}=a\end{matrix}\right.\)
Y gồm 0,3 - 2a mol SO2 dư
Y gồm 0,3 - 2a mol O2 dư
Y gồm 2a mol SO3
\(\Rightarrow0,3-2a+0,3-a+2a=0,5\)
\(\Leftrightarrow a=0,1\)
\(\Rightarrow\left\{{}\begin{matrix}n_{SO2\left(dư\right)}=0,1\left(mol\right)\\n_{O2\left(dư\right)}=0,2\left(mol\right)\\n_{SO3}=0,2\left(mol\right)\end{matrix}\right.\)
\(\%_{O2}=\%_{SO3}=\frac{0,2.100}{0,5}=40\%\)
\(\Rightarrow\%_{SO2}=100\%-40\%-40\%=20\%\)
b,\(H=\frac{0,1.100}{0,3}=33,33\%\)
