Đáp án:

 3)

a) 18g

b) 7,82 g

Giải thích các bước giải:

\(\begin{array}{l}
2)\\
{C_{12}}{H_{22}}{O_{11}} + {H_2}O \xrightarrow{xt} {C_6}{H_{12}}{O_6} + {C_6}{H_{12}}{O_6}\\
{C_6}{H_{12}}{O_6} \xrightarrow{ lên men} 2{C_2}{H_5}OH + 2C{O_2}\\
{C_2}{H_5}OH + {O_2} \xrightarrow{ lên men} C{H_3}COOH + {H_2}O\\
2C{H_3}COOH + 2K \to 2C{H_3}COOK + {H_2}\\
C{H_3}COOH + {C_2}{H_5}OH \xrightarrow{xt} C{H_3}COO{C_2}{H_5} + {H_2}O\\
C{H_3}COO{C_2}{H_5} + NaOH \to C{H_3}COONa + {C_2}{H_5}OH\\
3)\\
{C_{12}}{H_{22}}{O_{11}} + {H_2}O \xrightarrow{xt} {C_6}{H_{12}}{O_6} + {C_6}{H_{12}}{O_6}\\
a)\\
n{C_{12}}{H_{22}}{O_{11}} = \frac{m}{M} = \frac{{34,2}}{{342}} = 0,1\,mol\\
n{C_6}{H_{12}}{O_6} = n{C_{12}}{H_{22}}{O_{11}} = 0,1\,mol\\
m{C_6}{H_{12}}{O_6} = n \times M = 0,1 \times 180 = 18g\\
b)\\
{C_6}{H_{12}}{O_6} \xrightarrow{lên men} 2{C_2}{H_5}OH + 2C{O_2}\\
n{C_2}{H_5}OH = 2n{C_6}{H_{12}}{O_6} = 0,2\,mol\\
H = 85\%  \Rightarrow m{C_2}{H_5}OH = 0,2 \times 46 \times 85\%  = 7,82g
\end{array}\)