Đáp án:
\( {V_{{H_2}}} = 11,448{\text{ lít}}\)
\( {m_{C{H_3}COOH}} = 40,8{\text{ gam}}\)
\( {m_{giấm}} = 816{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2{C_2}{H_5}OH + 2Na\xrightarrow{{}}2{C_2}{H_5}ONa + {H_2}\)
\(2{H_2}O + 2Na\xrightarrow{{}}2NaOH + {H_2}\)
Ta có:
\({V_{{C_2}{H_5}OH}} = 50.92\% = 46{\text{ ml}} \to {{\text{V}}_{{H_2}O}} = 4{\text{ ml}}\)
\( \to {m_{{C_2}{H_5}OH}} = 46.0,8 = 36,8{\text{ gam;}}{{\text{m}}_{{H_2}O}} = 4.1 = 4{\text{ gam}}\)
\( \to {n_{{C_2}{H_5}OH}} = \frac{{36,8}}{{46}} = 0,8{\text{ mol;}}{{\text{n}}_{{H_2}O}} = \frac{4}{{18}} = \frac{2}{9}\)
\( \to {n_{{H_2}}} = \frac{1}{2}{n_{{C_2}{H_5}OH}} + \frac{1}{2}{n_{{H_2}O}} = \frac{{23}}{{45}}{\text{ mol}}\)
\( \to {V_{{H_2}}} = \frac{{23}}{{45}}.22,4 = 11,448{\text{ lít}}\)
\({C_2}{H_5}OH + {O_2}\xrightarrow{{men}}C{H_3}COOH + {H_2}O\)
\( \to {n_{C{H_3}COOH{\text{ lt}}}} = {n_{{C_2}{H_5}OH}} = 0,8\)
\( \to {n_{C{H_3}COOH}} = 0,8.85\% = 0,68{\text{ mol}}\)
\( \to {m_{C{H_3}COOH}} = 0,68.60 = 40,8{\text{ gam}}\)
\( \to {m_{giấm}} = \frac{{40,8}}{{5\% }} = 816{\text{ gam}}\)
