Đáp án:
a) ${C_3}{H_8}O(54,72%)$;${C_4}{H_{10}}O(45,28%)$
b) mCO2=14,96 g; mH2O=7,92 g
Giải thích các bước giải:
a) CTTQ 2 anken là ${C_{\overline n }}{H_{2\overline n }}$
${\overline M _{anken}} = 23,8.2 = 47,6$
$ \Rightarrow 14\overline n = 47,6 \Rightarrow \overline n = 3,4$
⇒ 2 anken là ${C_3}{H_6};{C_4}{H_8}$
⇒ 2 ancol là ${C_3}{H_8}O;{C_4}{H_{10}}O$
CTCT: $C{H_3} - C{H_2} - C{H_2}OH;C{H_3} - C{H_2} - C{H_2} - C{H_2}OH$
Áp dụng phương pháp đường chéo:
$\begin{array}{*{20}{c}}
{{C_3}{H_6}O:3}&{}&{}&{}&{0,6} \\
{}& \searrow &{}& \nearrow &{} \\
{}&{}&{3,4}&{}&{} \\
{}& \nearrow &{}& \searrow &{} \\
{{C_4}{H_8}O:4}&{}&{}&{}&{0,4}
\end{array} \Rightarrow \dfrac{{{n_{{C_3}{H_6}O}}}}{{{n_{{C_4}{H_8}O}}}} = \dfrac{3}{2}$
$\begin{gathered}
{n_{{C_3}{H_6}O}} = 3{\text{ mol; }}{n_{{C_4}{H_8}O}}{\text{ = 2 mol}} \hfill \\
\Rightarrow {{\text{m}}_{hh{\text{ ancol}}}} = 58.3 + 2.72 = 318\left( g \right) \hfill \\
\end{gathered} $
$\begin{gathered}
\Rightarrow \% {m_{{C_3}{H_6}O}} = \dfrac{{58.3}}{{318}}.100\% = 54,72\% \hfill \\
\Rightarrow \% {m_{{C_4}{H_8}O}} = 100 - 54,72 = 45,28\% \hfill \\
\end{gathered} $
b) CTTQ 2 ancol: ${C_{3,4}}{H_{8,8}}O$
$\begin{gathered}
{n_{ancol}} = \dfrac{{6,56}}{{65,6}} = 0,1{\text{ mol}} \hfill \\
\Rightarrow {{\text{n}}_{C{O_2}}} = 3,4{n_{ancol}} = 0,34mol \Rightarrow {m_{C{O_2}}} = 0,34.44 = 14,96g \hfill \\
\Rightarrow {n_{{H_2}O}} = 4,4{n_{ancol}} = 0,44mol \Rightarrow {m_{{H_2}O}} = 0,44.18 = 7,92g \hfill \\
\end{gathered} $
