Đáp án:
\( {m_{Na}} = 5,29{\text{ gam}}\)
\( {V_{{H_2}}} = 2,576{\text{ lít}}\)
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:
\({V_{{C_2}{H_5}OH}} = 11,25.92\% = 10,35{\text{ ml;}}{{\text{V}}_{{H_2}O}} = 0,9{\text{ ml}}\)
\( \to {m_{{C_2}{H_5}OH}} = 10,35.0,8 = 8,28{\text{ gam;}}{{\text{m}}_{{H_2}O}} = 0,9.1 = 0,9{\text{ gam}}\)
\( \to {n_{{C_2}{H_5}OH}} = \frac{{8,28}}{{46}} = 0,18{\text{ mol;}}{{\text{n}}_{{H_2}O}} = \frac{{0,9}}{{18}} = 0,05{\text{ mol}}\)
\( \to {n_{Na}} = {n_{{C_2}{H_5}OH}} + {n_{{H_2}O}} = 0,18 + 0,05 = 0,23{\text{ mol}}\)
\( \to {n_{{H_2}}} = \frac{1}{2}{n_{Na}} = 0,115{\text{ mol}}\)
\( \to {m_{Na}} = 0,23.23 = 5,29{\text{ gam}}\)
\( \to {V_{{H_2}}} = 0,115.22,4 = 2,576{\text{ lít}}\)
