Đáp án:
a) ${C_2}{H_6}O;{C_3}{H_8}O$
b) C2H6O: 43,4%; C3H8O: 56,6%
Giải thích các bước giải:
a) Gọi CTTQ 2 ancol là: ${C_n}{H_{2n + 2}}O$
${n_{{H_2}O}} = \dfrac{{0,9}}{{18}} = 0,05mol$
Bảo toàn nguyên tố $H$: ${n_{{H_2}}} = {n_{{H_2}O}} = 0,05mol \Rightarrow {n_{ancol}} = 2{n_{{H_2}}} = 0,1mol$
$ \Rightarrow {M_{ancol}} = \dfrac{{5,3}}{{0,1}} = 53 \Leftrightarrow 14n + 18 = 53 \Rightarrow n = 2,5$
CTPT 2 ancol là ${C_2}{H_6}O;{C_3}{H_8}O$
CTCT: $C{H_3} - C{H_2}OH;C{H_3} - C{H_2} - C{H_2}OH;C{H_3} - CH(OH) - C{H_3}$
b) Do n = 2,5
$\begin{gathered}
\Rightarrow {n_{{C_2}{H_6}O}} = {n_{{C_3}{H_8}O}} = \dfrac{{0,1}}{2} = 0,05mol \hfill \\
\Rightarrow \% {m_{{C_2}{H_6}O}} = \dfrac{{0,05.46}}{{5,3}}.100\% = 43,4\% \hfill \\
\Rightarrow \% {m_{{C_3}{H_8}O}} = 100 - 43,4\% = 56,6\% \hfill \\
\end{gathered} $