Đáp án + Giải thích các bước giải:

$V_{C_2H_5OH}=1380.8\%=110,4(ml)$

$n_{C_2H_5OH}=\dfrac{110,4.0,8}{46}=1,92(mol)$

$m_{dd C_2H_5OH}=m_{C_2H_5OH}+m_{H_2O}=110,4.0,8+(1380-110,4).1=1357,92(g)$

Phản ứng xảy ra:

$C_2H_5OH+O_2\xrightarrow{\text{Men}}CH_3COOH+H_2O$

Do $H=32\% \to n_{CH_3COOH}=n_{O_2}=1,92.32\%=0,6144(mol)$

$m_{\text{dd sau pứ}}=m_{dd C_2H_5OH}+m_{O_2}=1357,92+0,6144.32=1377,5808(g)$

$C\%_{CH_3COOH}=\dfrac{0,6144.60}{1377,5808}.100\%=2,676\%$