\(n_{C_2H_5OH\left(bđ\right)}=\dfrac{13,8}{46}=0,3\left(mol\right)\\ n_{CH_3COOH}=\dfrac{12}{60}=0,2\left(mol\right)\\ PTHH:C_2H_5OH+O_2\underrightarrow{\text{men giấm}}CH_3COOH\)
\(Theo.pt:n_{C_2H_5OH\left(pư\right)}=n_{CH_3COOH}=0,2\left(mol\right)\\ \rightarrow H=\dfrac{0,2}{0,3}=66,67\%\)