Đáp án:

 Câu 1: $C{H_3}CHO$; $\% {m_{C{H_3}OH}} = 25,8\% ;\% {m_{C{H_3}CHO}} = 74,2\% $

Câu 2: ${C_2}{H_5}OH$ ; ${C_3}{H_7}OH$ 

$\% {m_{{C_2}{H_5}OH}} = 60,53\% ;\% {m_{{C_3}{H_7}OH}} = 39,47\% $

Giải thích các bước giải:

 Câu 1: 

Gọi andehit là ${C_n}{H_{2n}}O$

${n_{{C_n}{H_{2n}}O}} = {n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2mol$

$ \Rightarrow {n_{{C_n}{H_{2n + 2}}O}} = {n_{{C_n}{H_{2n}}O}} = 0,2mol$

Y gồm $C{H_3}OH;{C_n}{H_{2n + 2}}O$

${n_{{H_2}}} = \dfrac{1}{2}{n_{C{H_3}OH}} + \dfrac{1}{2}{n_{{C_n}{H_{2n + 2}}O}} \Rightarrow {n_{C{H_3}OH}} = 0,15.2 - 0,2 = 0,1mol$

$ \Rightarrow {m_{C{H_3}OH}} = 0,1.32 = 3,2g \Rightarrow {m_{{C_n}{H_{2n}}O}} = 12,4 - 3,2 = 9,2g$

$ \Rightarrow {M_{{C_n}{H_{2n}}O}} = \dfrac{{9,2}}{{0,2}} = 46 \Rightarrow 14n + 16 = 46 \to n \approx 2$

⇒ andehit là $C{H_3}CHO$

$\begin{gathered}
  \% {m_{C{H_3}OH}} = \dfrac{{3,2}}{{12,4}}.100\%  = 25,8\%  \hfill \\
   \Rightarrow \% {m_{C{H_3}CHO}} = 100 - 25,8 = 74,2\%  \hfill \\ 
\end{gathered} $

Câu 2: Gọi CTTQ của 2 ancol là ${C_n}{H_{2n + 2}}O$

Đặt ${n_{{C_n}{H_{2n + 2}}O}} = x$

$ \Rightarrow {n_{Na}} = x;{n_{{H_2}}} = 0,5x$

Ta có: 

${m_{c.ran}} - {m_{ancol}} = {m_{Na}} - {m_{{H_2}}} \Rightarrow 23x - 0,5x.2 = 2,18 - 1,52 = 0,66$

$ \Rightarrow x = 0,03$

$ \Rightarrow {M_{ancol}} = \dfrac{{1,52}}{{0,03}} = 50,667 \Rightarrow 14n + 18 = 50,667 \Rightarrow n = 2,33$

⇒ 2 ancol là ${C_2}{H_5}OH$ (a mol); ${C_3}{H_7}OH$ (b mol)

ta có hpt: $\left\{ \begin{gathered}
  a + b = 0,03 \hfill \\
  46a + 60b = 1,52 \hfill \\ 
\end{gathered}  \right. \Rightarrow \left\{ \begin{gathered}
  a = 0,02 \hfill \\
  b = 0,01 \hfill \\ 
\end{gathered}  \right.$

$\begin{gathered}
   \Rightarrow \% {m_{{C_2}{H_5}OH}} = \frac{{0,02.46}}{{1,52}}.100\%  = 60,53\%  \hfill \\
   \Rightarrow \% {m_{{C_3}{H_7}OH}} = 100 - 60,53 = 39,47\%  \hfill \\ 
\end{gathered} $