Đáp án:
a/
$C_2H_4 + H_2O \xrightarrow{H^+} C_2H_5OH$
$C_2H_5OH + O_2 \xrightarrow{\text{men giấm}} CH_3COOH + H_2O$
$CH_3COOH + C_2H_5OH \overset{H_2SO_4}\leftrightarrows CH_3COOC_2H_5 + H_2O$
$CH_3COOC_2H_5 + NaOH → CH_3COONa + C_2H_5OH$
b/
$C_2H_4 + H_2O \xrightarrow{H^+} C_2H_5OH$
$C_2H_5OH + O_2 \xrightarrow{\text{men giấm}} CH_3COOH + H_2O$
$CH_3COOH + C_2H_5OH \overset{H_2SO_4}\leftrightarrows CH_3COOC_2H_5 + H_2O$
$C_2H_5OH + K → C_2H_5OK + \dfrac 12 H_2 \uparrow$
$CH_3COOH + KOH → CH_3COOK + H_2O$