Câu 3.
\(n_{CO_2}=\dfrac{33,6}{22,4}=1,5mol\)
Gọi \(\left\{{}\begin{matrix}n_{CH_4}=x\\n_{C_2H_4}=y\end{matrix}\right.\)
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)
x x ( mol )
\(C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\)
y 2y ( mol )
Ta có:
\(\left\{{}\begin{matrix}22,4x+22,4y=22,4\\x+2y=1,5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,5\\y=0,5\end{matrix}\right.\)
\(n_{hh}=\dfrac{22,4}{22,4}=1mol\)
\(\%V_{CH_4}=\dfrac{0,5}{1}.100=50\%\)
\(\%V_{C_2H_4}=\dfrac{0,5}{1}.100=50\%\)
Câu 2.
\(n_{CO_2}=\dfrac{8,96}{22,4}=0,4mol\)
\(n_{hh}=\dfrac{6,72}{22,4}=0,3mol\)
Gọi \(\left\{{}\begin{matrix}n_{CH_4}=x\\n_{C_2H_4}=y\end{matrix}\right.\)
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)
x x ( mol )
\(C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\)
y 2y ( mol )
Ta có:
\(\left\{{}\begin{matrix}x+y=0,3\\x+2y=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
\(\%V_{CH_4}=\dfrac{0,2}{0,3}.100=66,66\%\)
\(\%V_{C_2H_4}=100\%-66,66\%=33,34\%\)
úi , chụp có độ sáng zới , nhìn vậy hỏng mắt mất
