Coi \(m_{CH_4} = m_{C_2H_4} = 224(gam)\\ \Rightarrow n_{CH_4} = \dfrac{224}{16} = 14(mol)\\ \Rightarrow n_{C_2H_4} = \dfrac{224}{28} = 8(mol)\)
Vậy :
\(\%n_{CH_4} = \dfrac{14}{14+8}.100\% = 63,64\%\\ \%n_{C_2H_4} = 100\% - 63,64\% = 36,36\%\)
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\(GS:\)
\(n_{hh}=1\left(mol\right)\)
\(n_{CH_4}=a\left(mol\right),n_{C_2H_4}=b\left(mol\right)\)
\(\Rightarrow a+b=1\left(1\right)\)
\(TC:\)
\(16a=28b\left(2\right)\)
\(\left(1\right),\left(2\right):a=\dfrac{7}{11},b=\dfrac{4}{11}\)
\(\%n_{CH_4}=\dfrac{7}{11}\cdot100\%=63.64\%\)
\(\%n_{C_2H_4}=36.36\%\)
Em xem thử cách làm này nhé !!
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