Đáp án:

$43,2(g)$

Giải thích các bước giải:

$V_{C_2H_5OH}=\frac{287,5.16}{100}=46(ml)$

$\to m_{C_2H_5OH}=46.0,8=36,8(g)$

$\to n_{C_2H_5OH(lt)}=\frac{36,8}{46}=0,8(mol)$

$\to n_{C_2H_5OH(tt)}=0,8.90\%=0,72(mol)$

$C_2H_5OH+O_2\xrightarrow{\rm men\,giấm}CH_3COOH+H_2O$

Theo PT: $n_{CH_3COOH(tt)}=n_{C_2H_5OH(tt)}=0,72(mol)$

$\to m_{CH_3COOH(tt)}=0,72.60=43,2(g)$

Đáp án:

\( {m_{C{H_3}COOH}} = 43,2{\text{ gam}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\({C_2}{H_5}OH + {O_2}\xrightarrow{{men}}C{H_3}COOH + {H_2}O\)

Ta có:

\({V_{{C_2}{H_5}OH}} = 287,5.16\%  = 46{\text{ ml}}\)

\( \to {m_{{C_2}{H_5}OH}} = 46.0,8 = 36,8{\text{ gam}}\)

\( \to {n_{{C_2}{H_5}OH}} = \frac{{36,8}}{{46}} = 0,8{\text{ mol = }}{{\text{n}}_{C{H_3}COOH{\text{ lt}}}}\)

\( \to {m_{C{H_3}COOH}} = 0,8.90\%  = 0,72{\text{ mol}}\)

\( \to {m_{C{H_3}COOH}} = 0,72.60 = 43,2{\text{ gam}}\)