a) Gọi \(\left\{{}\begin{matrix}n_{C_2H_4}=a\left(mol\right)\\n_{C_3H_6}=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\Rightarrow a+b=\dfrac{6,72}{22,4}=0,3\left(1\right)\)
PTHH:
\(C_2H_4+3O_2\xrightarrow[]{t^o}2CO_2+2H_2O\)
a-------->3a------>2a
\(2C_3H_6+9O_2\xrightarrow[]{t^o}6CO_2+6H_2O\)
b-------->4,5b---->3b
\(\Rightarrow n_{O_2}=3a+4,5b=\dfrac{23,52}{22,4}=1,05\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,2}{0,3}.100\%=66,67\%\\\%V_{C_3H_6}=100\%-66,67\%=33,33\%\end{matrix}\right.\)
b) \(V_{CO_2}=\left(0,2.2+0,1.3\right).22,4=15,68\left(l\right)\)
