Đáp án:
b) 0,15M
c) 6,9 ml
Giải thích các bước giải:
a) $C{H_2}OH - {\left( {CHOH} \right)_4} - CHO + 2AgN{O_3} + 3N{H_3} + {H_2}O \to C{H_2}OH - {\left( {CHOH} \right)_4} - COON{H_4} + 2Ag + 2N{H_4}N{O_3}$
b)
$\begin{gathered}
{n_{Ag}} = \dfrac{{16,2}}{{108}} = 0,15mol \hfill \\
\Rightarrow {n_{{C_6}{H_{12}}{O_6}}} = \dfrac{1}{2}{n_{Ag}} = 0,075mol \hfill \\
\Rightarrow {C_M} = \dfrac{n}{V} = \dfrac{{0,075}}{{0,5}} = 0,15M \hfill \\
\end{gathered} $
c)
${C_6}{H_{12}}{O_6}\xrightarrow{{men}}2{C_2}{H_5}OH + 2C{O_2}$
$\begin{gathered}
{\text{Theo PTHH:}}{\text{ }}{n_{{C_2}{H_5}OH}} = 2{n_{{C_6}{H_{12}}{O_6}}} = 0,15mol \hfill \\
\Rightarrow {m_{{C_2}{H_5}OH}} = 0,15.46 = 6,9g \hfill \\
H = 80\% \Rightarrow {m_{{C_2}{H_5}OH(tt)}} = \dfrac{{6,9.80}}{{100}} = 5,52g \hfill \\
\Rightarrow {V_{{C_2}{H_5}OH}} = \frac{m}{D} = \dfrac{{5,52}}{{0,8}} = 6,9ml \hfill \\
\end{gathered} $
