Đáp án:
\( {V_{{H_2}}} = 18,288{\text{ lít}}\)
\({m_{Na}} = 37,56{\text{ gam}}\)
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:
\({V_{{C_2}{H_5}OH}} = 50.60\% = 30{\text{ ml;}}{{\text{V}}_{{H_2}O}} = 50 - 30 = 20{\text{ ml}}\)
\( \to {m_{{C_2}{H_5}OH}} = 30.0,8 = 24{\text{ gam;}}{{\text{m}}_{{H_2}O}} = 20.1 = 20{\text{ gam}}\)
\( \to {n_{{C_2}{H_5}OH}} = \frac{{24}}{{46}} = \frac{{12}}{{23}};{n_{{H_2}O}} = \frac{{20}}{{18}} = \frac{{10}}{9}\)
\({n_{Na}} = {n_{{C_2}{H_5}OH}} + {n_{{H_2}O}} = \frac{{12}}{{23}} + \frac{{10}}{9} = \frac{{338}}{{207}}\)
\({n_{{H_2}}} = \frac{1}{2}{n_{Na}} = \frac{{169}}{{207}}\)
\( \to {V_{{H_2}}} = \frac{{169}}{{207}}.22,4 = 18,288{\text{ lít}}\)
\({m_{Na}} = \frac{{338}}{{207}}.23 = 37,56{\text{ gam}}\)
