Đáp án:
Câu 9: $\% {m_{glixerol}} = 40\% ;\% {m_{e\tan ol}} = 60\% $
Câu 10: $\% {m_{C{H_3}CHO}} = 20,18\% ;\% {m_{{C_2}{H_5}CHO}} = 79,82\% $
Giải thích các bước giải:
Câu 9:
${n_{Cu{{(OH)}_2}}} = \dfrac{{4,9}}{{98}} = 0,05mol;{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3mol$
$Cu{(OH)_2} + 2{C_3}{H_5}{(OH)_3} \to {\left[ {{C_3}{H_5}{{(OH)}_2}O} \right]_2}Cu + 2{H_2}O$
$\begin{gathered}
{C_3}{H_5}{(OH)_3} + 3Na \to {C_3}{H_5}{(ONa)_3} + \frac{3}{2}{H_2} \hfill \\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2} \hfill \\
\end{gathered} $
Theo PTHH: ${n_{glixerol}} = 2{n_{Cu{{(OH)}_2}}} = 0,05.2 = 0,1mol$
$2{n_{{H_2}}} = 3{n_{glixerol}} + {n_{e\tan ol}} \Rightarrow {n_{e\tan ol}} = 2.0,3 - 3.0,1 = 0,3mol$
$ \Rightarrow {m_{glixerol}} = 0,1.92 = 9,2g;{m_{e\tan ol}} = 0,3.46 = 13,8g$
$\begin{gathered}
\Rightarrow \% {m_{glixerol}} = \dfrac{{9,2}}{{9,2 + 13,8}}.100\% = 40\% \hfill \\
\Rightarrow \% {m_{e\tan ol}} = 100 - 40 = 60\% \hfill \\
\end{gathered} $
Câu 10:
${n_{Ag}} = \dfrac{{43,2}}{{108}} = 0,4mol$
$\begin{gathered}
C{H_3}CHO + 2AgN{O_3} + 3N{H_3} + {H_2}O \to C{H_3}COON{H_4} + 2Ag + 2N{H_4}N{O_3} \hfill \\
{C_2}{H_5}CHO + 2AgN{O_3} + 3N{H_3} + {H_2}O \to {C_2}{H_5}COON{H_4} + 2Ag + 2N{H_4}N{O_3} \hfill \\
\end{gathered} $
Gọi x, y là số mol của $C{H_3}CHO$; ${C_2}{H_5}CHO$
Ta có hpt: $\left\{ \begin{gathered}
2x + 2y = 0,4 \hfill \\
77x + 91y = 17,5 \hfill \\
\end{gathered} \right. \Rightarrow \left\{ \begin{gathered}
x = 0,05 \hfill \\
y = 0,15 \hfill \\
\end{gathered} \right.$
$ \Rightarrow {m_{C{H_3}CHO}} = 0,05.44 = 2,2g;{m_{{C_2}{H_5}CHO}} = 0,15.58 = 8,7g$
$\begin{gathered}
\Rightarrow \% {m_{C{H_3}CHO}} = \dfrac{{2,2}}{{2,2 + 8,7}}.100\% = 20,18\% \hfill \\
\Rightarrow \% {m_{{C_2}{H_5}CHO}} = 100 - 20,18 = 79,82\% \hfill \\
\end{gathered} $
