Đáp án:

 281,74 g

Giải thích các bước giải:

\(\begin{array}{l}
V{C_2}{H_5}OH = \dfrac{{V{\rm{dd}} \times Dr}}{{100}} = \dfrac{{2000 \times 15}}{{100}} = 300g\\
m{C_2}{H_5}OH = V \times D = 0,8 \times 300 = 240g\\
n{C_2}{H_5}OH = \dfrac{m}{M} = \dfrac{{240}}{{46}} = \dfrac{{120}}{{23}}\,mol\\
{C_2}{H_5}OH + {O_2} \to C{H_3}COOH + {H_2}O\\
nC{H_3}COOH = n{C_2}{H_5}OH = \dfrac{{120}}{{23}}\,mol\\
H = 90\%  \Rightarrow mC{H_3}COOH = \dfrac{{120}}{{23}} \times 90\%  \times 60 = 281,74g
\end{array}\)

$V_{C_2H_5OH}=\frac{V_{hh}.D_r}{100}=\frac{2.15}{100}=0,3(l)=300(ml)$

$m_{C_2H_5OH}=300.0,8=240(g)$

$n_{C_2H_5OH}=240:46.90%=4,7(mol)$

$PTHH:C_2H_5OH+O_2\xrightarrow{men giấm}CH_3COOH+H_2O$

$(mol)--4,7-------4,7--$

$m_{CH_3COOH}=4,7.60=282(g)$