Đáp án:

$C\%_{CH_3COOH}\approx 7,99\%$

Giải thích các bước giải:

Trong $500ml$ rượu có:

$+)V_{C_2H_5OH}=\dfrac{500.10}{100}=50(ml)$

$\Rightarrow m_{C_2H_5OH}=50.0,78=39(g)$

$\Rightarrow n_{C_2H_5OH}=\dfrac{39}{46}\approx 0,85(mol)$

$+)V_{H_2O}=500-50=450(ml)$

$\Rightarrow m_{H_2O}=450.1=450(g)$

$CH_3CH_2OH+O_2\to CH_3COOH+H_2O\\ 0,85\;\;\;\longrightarrow\;\;\;\;\;\; 0,85\;\;\;\longrightarrow \;0,85(mol)$

Vì $H\%=80\%$ nên $m_{CH_3COOH}=0,85.60.80\%=40,8(g)$

Bảo toàn khối lượng:

$m_{ddrượu}+m_{O_2}=m_{\text{dd sau}}$

$\Rightarrow m_{\text{dd sau}}=39+450+0,85.32.80\%=510,76(g)$

$\Rightarrow C\%_{CH_3COOH}=\dfrac{40,8}{510,76}.100\%\approx 7,99\%$

V(C2H5OH)=500*10/100=50ml=>v H2O=500-50=450ml

=>m(C2H5OH)=50*0,78=39g

m H2O=450*1=450g

n(C2H5OH)=39/46=mol

n CH3COOH=nO2=n(C2H5OH)*H=39/46*80%=78/115 mol

mdd sau pư=mO2+m(C2H5OH)+m H2O=78/115*32+39+450=510,7g

m( CH3COOH)=78/115*60=40,7g

=>C%( CH3COOH)=40,7/510,7*100%=7,97%