a,
$V_{C_2H_5OH}= 3,45.10\%=0,345l=345ml$
$\Rightarrow m_{C_2H_5OH}=345.0,8=276g$
$\Rightarrow n_{C_2H_5OH}=\frac{276}{46}= 6 mol$
$C_2H_5OH+O_2\buildrel{{\text{men giấm}}}\over\to CH_3COOH+H_2O$
Theo lí thuyết tạo 6 mol axit.
$H=90\%$
$\Rightarrow m_{CH_3COOH}=6.60.90\%=324g$
b,
$m_{\text{giấm}}=324:5\%=6480g$
