Đáp án:
\( {m_{{C_2}{H_5}OH}} = 73,6{\text{ gam}}\)
\( {m_{{C_6}{H_{12}}{O_6}}} = 160{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\({C_6}{H_{12}}{O_6}\xrightarrow{{men,90\% }}2{C_2}{H_5}OH + 2C{O_2}\)
Ta có:
\({n_{C{O_2}}} = \frac{{35,84}}{{44}} = 1,6{\text{ mol = }}{{\text{n}}_{{C_2}{H_5}OH}}\)
\( \to {m_{{C_2}{H_5}OH}} = 1,6.46 = 73,6{\text{ gam}}\)
Ta có:
\({n_{{C_6}{H_{12}}{O_6}{\text{ lt}}}} = \frac{1}{2}{n_{C{O_2}}} = 0,8{\text{ mol}}\)
\( \to {n_{{C_6}{H_{12}}{O_6}}} = \frac{{0,8}}{{90\% }} = \frac{8}{9}{\text{ mol}}\)
\( \to {m_{{C_6}{H_{12}}{O_6}}} = \frac{8}{9}.(12.6 + 12 + 16.6) = 160{\text{ gam}}\)