Đáp án:
\(\begin{array}{l}
a)\\
{m_{{C_2}{H_5}OH}} = 13,8g\\
{V_{{C_2}{H_5}OH}} = 17,25\,ml\\
b)\\
{m_{{C_6}{H_{12}}{O_6}}} = 28,42g\\
c)\\
{m_{{\rm{dd}}C{H_3}COOH}} = 405g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{C_6}{H_{12}}{O_6} \xrightarrow{\text{ lên men }} 2{C_2}{H_5}OH + 2C{O_2}\\
{n_{C{O_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
{n_{{C_2}{H_5}OH}} = {n_{C{O_2}}} = 0,3\,mol\\
{m_{{C_2}{H_5}OH}} = 0,3 \times 46 = 13,8g\\
{V_{{C_2}{H_5}OH}} = \dfrac{{13,8}}{{0,8}} = 17,25\,ml\\
b)\\
{n_{{C_6}{H_{12}}{O_6}}} = \dfrac{{0,3}}{2} = 0,15\,mol\\
H = 95\% \Rightarrow {m_{{C_6}{H_{12}}{O_6}}} = \dfrac{{0,15 \times 180 \times 100}}{{95}} = 28,42g\\
c)\\
{C_2}{H_5}OH + {O_2} \xrightarrow{\text{ lên men }} C{H_3}COOH + {H_2}O\\
{n_{C{H_3}COOH}} = {n_{{C_2}{H_5}OH}} = 0,3\,mol\\
H = 90\% \Rightarrow {m_{C{H_3}COOH}} = 0,3 \times 60 \times 90\% = 16,2g\\
{m_{{\rm{dd}}C{H_3}COOH}} = \dfrac{{16,2 \times 100}}{4} = 405g
\end{array}\)
