a, \(2K+2CH_3COOH\rightarrow2CH_3COOK+H_2\)
b, \(C_2H_5OH+O_2\underrightarrow{t^o,xt}CH_3COOH+H_2O\)
Ta có: \(n_{CH_3COOH}=\dfrac{12}{60}=0,2\left(mol\right)\)
Theo PT: \(n_{C_2H_5OH\left(LT\right)}=n_{CH_3COOH}=0,2\left(mol\right)\)
Mà: H = 80% \(\Rightarrow n_{C_2H_5OH\left(TT\right)}=\dfrac{0,2}{80\%}=0,25\left(mol\right)\)
\(\Rightarrow m_{C_2H_5OH\left(TT\right)}=0,25.46=11,5\left(g\right)\)
\(\Rightarrow V_{C_2H_5OH\left(TT\right)}=\dfrac{11,5}{0,8}=14,375\left(ml\right)\)
\(\Rightarrow V_{C_2H_5OH\left(10^o\right)}=\dfrac{14,375}{10}.100=143,75\left(ml\right)\)