\(n_{h^2khí}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ m_{tăng}=m_{C_2H_4}=2,8\left(g\right)\Rightarrow n_{C_2H_4}=\dfrac{2,8}{28}=0,1\left(mol\right)\\ n_{CH_4}=0,3-0,1=0,2\left(mol\right)\)
a) \(V_{CH_4}=0,2.22,4=4,48\left(l\right)\Rightarrow V_{C_2H_4}=2,24\left(l\right)\)
b) \(\%V_{CH_4}=\%n_{CH_4}=\dfrac{0,2}{0,3}\approx0,67\left(\%\right)\approx67\left(\%\right)\\ \%C_2H_4=100-67=33\left(\%\right)\)
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