\(C_2H_5OH+O_2\rightarrow\left(men.giấm\right)CH_3COOH+H_2O\\ n_{CH_3COOH}=\dfrac{200.12\%}{60}=0,4\left(mol\right)\\ n_{C_2H_5OH}=n_{CH_3COOH}=0,4\left(mol\right)\\ m_{C_2H_5OH}=0,4.46=18,4\left(g\right)\\ V_{C_2H_5OH}=\dfrac{18,4}{0,8}=23\left(ml\right)\)
Ta có: \(m_{CH_3COOH}=200.12\%=24\left(g\right)\Rightarrow n_{CH_3COOH}=\dfrac{24}{60}=0,4\left(mol\right)\)
PT: \(C_2H_5OH+O_2\underrightarrow{^{mengiam}}CH_3COOH+H_2O\)
Theo PT: \(n_{C_2H_5OH}=n_{CH_3COOH}=0,4\left(mol\right)\)
\(\Rightarrow m_{C_2H_5OH}=0,4.46=18,4\left(g\right)\)
\(\Rightarrow V_{C_2H_5OH}=\dfrac{18,4}{0,8}=23\left(ml\right)\)
