Đáp án:
\(D\)
Giải thích các bước giải:
Sơ đồ phản ứng:
\({C_6}{H_{10}}{O_5}\xrightarrow{{70\% }}2{C_2}{H_5}OH\)
Ta có:
\({V_{{C_2}{H_5}OH}} = 200.40\% = 80{\text{ lít}}\)
\( \to {m_{{C_2}{H_5}OH}} = 80.0,8 = 64{\text{ kg}}\)
\( \to {n_{{C_2}{H_5}OH}} = \frac{{64}}{{46}} = \frac{{32}}{{23}}{\text{ kmol}}\)
\( \to {n_{{C_6}{H_{10}}{O_{5{\text{ lt}}}}}} = \frac{1}{2}{n_{{C_2}{H_5}OH}} = \frac{{16}}{{23}}{\text{ kmol}}\)
\( \to {n_{{C_6}{H_{10}}{O_5}}} = \frac{{\frac{{16}}{{23}}}}{{70\% }} = \frac{{160}}{{161}}{\text{ kmol}}\)
\( \to {m_{{C_6}{H_{10}}{O_5}}} = \frac{{160}}{{161}}.162 = 161{\text{ kg}}\)
\( \to {m_{mùn{\text{ cưa}}}} = \frac{{161}}{{50\% }} = 322{\text{ kg}}\)