\(C_2H_5OH+O_2\underrightarrow{lên-men-giâm}CH_3COOH+H_2O\)
\(n_{C_2H_5OH}=\dfrac{4,6}{46}=0,1\left(mol\right)\)
Theo PTHH: \(n_{CH_3COOH}\left(lt\right)=0,1\left(mol\right)\)
Vì \(H=80\%\)
\(\Rightarrow n_{CH_3COOH}\left(tt\right)=0,08\left(mol\right)\)
\(\Rightarrow m_{CH_3COOH}\left(tt\right)=4,8\left(g\right)\)