Đáp án:
\({m_{{C_6}{H_{10}}{O_5}}} = 9{\text{ kg}}\)
Giải thích các bước giải:
Sơ đồ phản ứng:
\({C_6}{H_{10}}{O_5}\xrightarrow{{72\% }}2{C_2}{H_5}OH\)
Ta có:
\({V_{{C_2}{H_5}OH}} = 10.46\% = 4,6{\text{ lít}}\)
\( \to {m_{{C_2}{H_5}OH}} = 4,6.0,8 = 3,68{\text{ kg}}\)
\( \to {n_{{C_2}{H_5}OH}} = \frac{{3,68}}{{46}} = 0,08{\text{ kmol}}\)
\( \to {n_{{C_6}{H_{10}}{O_5}{\text{ }}lt}} = \frac{1}{2}{n_{{C_2}{H_5}OH}} = 0,04{\text{ kmol}}\)
\( \to {m_{{C_6}{H_{10}}{O_5}{\text{ lt}}}} = 0,04.162 = 6,48{\text{ kg}}\)
\( \to {m_{{C_6}{H_{10}}{O_5}}} = \frac{{6,48}}{{72\% }} = 9{\text{ kg}}\)