Giúp em bài này với mn nhé

$V_{C_2H_5OH}= 4,6.14\%= 0,644l= 644ml$ 

=> $m_{C_2H_5OH}= 644.0,8= 515,2g$

=> $n_{C_2H_5OH}= \frac{515,2}{46}= 11,2 mol$

$V_{H_2O}= 4,6-0,644= 3,956l= 3956ml$

=> $m_{H_2O}= 3956g$ 

$H= 30\%$ nên $n_{C_2H_5OH pứ}= 11,2.30\%= 3,36 mol$

$C_2H_5OH+ O_2 \buildrel{{\text{men giấm}}}\over\to CH_3COOH+ H_2O$ 

=> $n_{O_2}= n_{CH_3COOH}= 3,36 mol$

$m_{dd spứ}= m_{C_2H_5OH bđ}+ m_{H_2O bđ}+ m_{O_2}= 515,2+3956+3,36.32= 4578,72g$

$C\%_{CH_3COOH}= \frac{3,36.60.100}{4578,72}= 4,4\%$