Đáp án:
\(C\)
Giải thích các bước giải:
\({C_6}{H_{12}}{O_6}\xrightarrow{{men}}2{C_2}{H_5}OH + 2C{O_2}\)
Ta có:
\({V_{{C_2}{H_5}OH}} = 50.90\% = 45{\text{ lít}}\)
\( \to {m_{{C_2}{H_5}OH}} = 45.0,8 = 36{\text{ kg}}\)
\( \to {n_{{C_2}{H_5}OH}} = \frac{{36}}{{46}} = \frac{{18}}{{23}}{\text{ kmol}}\)
\( \to {n_{{C_6}{H_{12}}{O_6}{\text{lt}}}} = \frac{1}{2}{n_{{C_2}{H_5}OH}} = \frac{9}{{23}}{\text{ kmol}}\)
\( \to {n_{{C_6}{H_{12}}{O_6}}} = \frac{{\frac{9}{{23}}}}{{80\% }} = \frac{{45}}{{92}}{\text{ kmol}}\)
\( \to {m_{{C_6}{H_{12}}{O_6}}} = \frac{{45}}{{92}}.180 = 88{\text{ kg = 0}}{\text{,088 tấn}}\)
\( \to \% {m_{{C_6}{H_{12}}{O_6}}} = \frac{{0,088}}{1} = 8,8\% \)