$CH_2=CH_2+H_2O\xrightarrow{170^oC,H_2SO_4\text{ loãng}}CH_3CH_2OH$
$1\xrightarrow{}1\xrightarrow{}1(mol)$
$n_{C_2H_4}=\dfrac{22,4}{22,4}=1(mol)$
$⇒n_{C_2H_5OH}\text{giả thuyết}=1(mol)$
$n_{C_2H_5OH}\text{thực tế}=\dfrac{13,8}{46}=0,3(mol)$
$H\%=\dfrac{n_{\text{thực tế}}}{n_{\text{giả thuyết}}}.100=\dfrac{0,3}{1}.100=30\%$