Đáp án:
\(\begin{array}{l}
a.{m_{{C_2}{H_5}OH}} = 9,2g\\
b.{m_{{C_6}{H_{12}}{O_6}(tt)}} = 20g\\
c.H = 45,45\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
{C_6}{H_{12}}{O_6} \to 2{C_2}{H_5}OH + 2C{O_2}\\
{n_{C{O_2}}} = 0,2mol\\
a.\\
{n_{{C_2}{H_5}OH}} = {n_{C{O_2}}} = 0,2mol\\
\to {m_{{C_2}{H_5}OH}} = 9,2g\\
b.\\
{n_{{C_6}{H_{12}}{O_6}}} = \dfrac{1}{2}{n_{C{O_2}}} = 0,1mol\\
\to {m_{{C_6}{H_{12}}{O_6}}} = 18g\\
\to {m_{{C_6}{H_{12}}{O_6}(tt)}} = \dfrac{{18 \times 100}}{{90}} = 20g
\end{array}\)
\(\begin{array}{l}
c.\\
{C_2}{H_5}OH + C{H_3}COOH \to C{H_3}COO{C_2}{H_5} + {H_2}O\\
{n_{C{H_3}COOH}} = 0,1mol\\
\to {n_{C{H_3}COOH}} < {n_{{C_2}{H_5}OH}} \to {n_{{C_2}{H_5}OH}}dư\\
\to {n_{C{H_3}COO{C_2}{H_5}}} = {n_{C{H_3}COOH}} = 0,1mol\\
\to {m_{C{H_3}COO{C_2}{H_5}}} = 8,8\\
\to H = \dfrac{4}{{8,8}} \times 100 = 45,45\%
\end{array}\)