`C_2H_5OH+O_2` $\xrightarrow[25-30^oC]{\text{men giấm}}$ `CH_3COOH+H_2O`

`Theo\ pt:n_{C_2H_5OH} = n_{CH_3COOH} = 12/60 = 0,2\ (mol)`

Ta có `:`

`H=m_{t t}/m_{l t} xx 100`

`<=> 60 = (0,2 xx 45)/m_{l t} xx 100`

`-> m_{l t} ~~ 15,33\ (g)`

`n_{CH_3COOH}=12/60 =0,2(mol)`

`PTHH: C_2H_5OH+O_2 \overset(25-30^oC , men giấm ) CH_3COOH+H_2O`

`n_{C_2H_5OH}=n_{CH_3COOH}=0,2(mol)`

`->m_{C_2H_5OH} thực tế=0,2.46=9,2(g)`

`H=60%=>m_{C_2H_5OH} l t=9,2 :60% ~~15,33(g)`