Đáp án:
$m_{CH_3CH_2OH}=4,6g\\m_{CH_3COOH}=6g$
Giải thích các bước giải:
1. $CH_2=CH_2+H_2)\xrightarrow{H^+,t^o,xt}CH_3CH_2OH\\CH_3CH_2OH+O_2\xrightarrow{men\ giấm}CH_3COOH+H_2O$
2. $n_{C_2H_4}=\dfrac{2,24}{22,4}=0,1\ mol$
$CH_2=CH_2+H_2O\xrightarrow{H^+,t^o,xt}CH_3CH_2OH\\0,1\hspace{6cm}0,1\\CH_3CH_2OH+O_2\xrightarrow{men\ giấm}CH_3COOH+H_2O\\0,1\hspace{6cm}0,1$
Vậy: $n_{CH_3CH_2OH}=0,1\ mol ⇒m_{CH_3CH_2OH}=0,1.46=4,6g\\n_{CH_3COOH}=0,1\ mol ⇒m_{CH_3COOH}=0,1.60=6g$