Đáp án:
\({m_{{C_6}{H_{12}}{O_6}}} = 18,89{\text{ gam}}\)
Giải thích các bước giải:
Quy tinh bột về gốc \(C_6H_{10}O_5\).
\( \to {m_{{C_6}{H_{10}}{O_5}}} = 1000.2\% = 20{\text{ gam}}\)
\({C_6}{H_{10}}{O_5} + {H_2}O\xrightarrow{{{H^ + }/H = 85\% }}{C_6}{H_{12}}{O_6}\)
Ta có:
\({n_{{C_6}{H_{10}}{O_5}}} = \frac{{20}}{{162}} = {n_{{C_6}{H_{12}}{O_6}{\text{ lt}}}}\)
\({n_{{C_6}{H_{12}}{O_6}}} = \frac{{20}}{{162}}.85\% = \frac{{17}}{{162}}{\text{ mol}}\)
\( \to {m_{{C_6}{H_{12}}{O_6}}} = \frac{17}{{162}}.180 = 18,89{\text{ gam}}\)
