Đáp án:
\({V_{{H_2}}} = 9,159{\text{ lít}}\)
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:
\({V_{{C_2}{H_5}OH}} = 40.92\% = 36,8{\text{ ml}} \to {{\text{m}}_{{H_2}O}} = 3,2{\text{ ml}}\)
\( \to {m_{{C_2}{H_5}OH}} = 36,8.0,8 = 29,44{\text{ gam}}\)
\({m_{{H_2}O}} = 3,2.1 = 3,2{\text{ gam}}\)
\( \to {n_{{C_2}{H_5}OH}} = \frac{{29,44}}{{46}} = 0,64{\text{ mol;}}{{\text{n}}_{{H_2}O}} = \frac{{3,2}}{{18}} = \frac{8}{{45}}{\text{ mol}}\)
\( \to {n_{{H_2}}} = \frac{1}{2}{n_{{C_2}{H_5}OH}} + \frac{1}{2}{n_{{H_2}O}} = \frac{{92}}{{225}}{\text{ mol}}\)
\( \to {V_{{H_2}}} = \frac{{92}}{{225}}.22,4 = 9,159{\text{ lít}}\)
