Đáp án:

\( {m_{{C_2}{H_5}OH}} = 9,2{\text{ gam}}\)

\({m_{{C_6}{H_{12}}{O_6}}} = 20{\text{ gam}}\)

\(H = 45,5\% \)

Giải thích các bước giải:

Phản ứng xảy ra:

\({C_6}{H_{12}}{O_6}\xrightarrow{{men}}2{C_2}{H_5}OH + 2C{O_2}\)

 Ta có:

\({n_{C{O_2}}} = \frac{{4,48}}{{22,4}} = 0,2{\text{ mol = }}{{\text{n}}_{{C_2}{H_5}OH}}\)

\( \to {m_{{C_2}{H_5}OH}} = 0,2.(12.2 + 5 + 17) = 9,2{\text{ gam}}\)

Ta có:

\({n_{{C_6}{H_{12}}{O_6}{\text{ lt}}}} = \frac{1}{2}{n_{C{O_2}}} = \frac{1}{2}.0,2 = 0,1{\text{ mol}}\)

\( \to {n_{{C_6}{H_{12}}{O_6}}} = \frac{{0,1}}{{90\% }} \to {m_{{C_6}{H_{12}}{O_6}}} = \frac{{0,1}}{{90\% }}.(12.6 + 12 + 16.6) = 20{\text{ gam}}\)

Cho lượng rượu trên đem este hóa

\(C{H_3}COOH + {C_2}{H_5}OH\xrightarrow{{{H_2}S{O_4},{t^o}}}C{H_3}COO{C_2}{H_5} + {H_2}O\)

Ta có:

\({n_{C{H_3}COOH}} = \frac{6}{{60}} = 0,1{\text{ mol < }}{{\text{n}}_{ancol}}\)

Vậy hiệu suất tính theo axit

\({n_{C{H_3}COO{C_2}{H_5}}} = \frac{4}{{12 + 3 + 12 + 16.2 + 12.2 + 5}} = \frac{4}{{88}} = \frac{1}{{22}}{\text{ mol = }}{{\text{n}}_{C{H_3}{\text{CO}}{\text{O}}{\text{H phản ứng}}}}\)

Hiệu suất \(H = \frac{{\frac{1}{{22}}}}{{0,1}} = 45,5\% \)