Đáp án:
\( {V_{dd{\text{ rượu}}}} = 2,875{\text{ lít}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\({C_6}{H_{12}}{O_6}\xrightarrow{{men}}2{C_2}{H_5}OH + 2C{O_2}\)
Ta có:
\({m_{{C_6}{H_{12}}{O_6}}} = 2,5.(100\% - 20\% ) = 2{\text{ kg}}\)
\( \to {n_{{C_6}{H_{12}}{O_6}}} = \frac{2}{{180}} = \frac{1}{{90}}{\text{ kmol}}\)
\( \to {n_{{C_2}{H_5}OH{\text{ lt}}}} = 2{n_{{C_6}{H_{12}}{O_6}}} = \frac{1}{{90}}.2 = \frac{2}{{90}}{\text{ kmol}}\)
Vì hao hụt 10%.
\( \to {n_{{C_2}{H_5}OH}} = \frac{2}{{90}}.(100\% - 10\% ) = 0,02{\text{ kmol}}\)
\( \to {m_{{C_2}{H_5}OH}} = 0,02.46 = 0,92{\text{ kg}}\)
\( \to {V_{{C_2}{H_5}OH}} = \frac{{0,92}}{{0,8}} = 1,15{\text{ lít}}\)
\( \to {V_{dd{\text{ rượu}}}} = \frac{{{V_{{C_2}{H_5}OH}}}}{{40\% }} = \frac{{1,15}}{{40\% }} = 2,875{\text{ lít}}\)
