Đáp án:
\( {m_{C{H_3}COOH}} = 333,91{\text{ gam}}\)
\( {m_{giấm}} = 6678,2{\text{ gam}}\)
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:
\({V_{{C_2}{H_5}OH}} = 5000.8\% = 400{\text{ ml}}\)
\( \to {m_{{C_2}{H_5}OH}} = 400.0,8 = 320{\text{ gam}}\)
\( \to {n_{{C_2}{H_5}OH}} = \frac{{320}}{{46}} = \frac{{160}}{{23}} = {n_{C{H_3}COOH{\text{ lt}}}}\)
\( \to {n_{C{H_3}COOH}} = \frac{{160}}{{23}}.80\% = \frac{{128}}{{23}}{\text{ mol}}\)
\( \to {m_{C{H_3}COOH}} = \frac{{128}}{{23}}.60 = 333,91{\text{ gam}}\)
\( \to {m_{giấm}} = \frac{{333,91}}{{5\% }} = 6678,2{\text{ gam}}\)
