Đáp án:

\(\begin{array}{l}
1)\quad \%m_{C_6H_5OH} = 74,6\%\\
2)\quad \rm X:\quad CH_3COOH\\
3)\quad \rm Y:\quad HOOC-COOH\\
4)\quad \rm A:\quad CH_2=CHCOOH
\end{array}\) 

Giải thích các bước giải:

\(\begin{array}{l}
1)\\
\rm C_6H_5OH + Na \longrightarrow C_6H_5ONa + \dfrac12H_2\uparrow\\
\rm CH_3OH + Na \longrightarrow CH_3ONa + \dfrac12H_2\uparrow\\
hh\ A: \ a\ mol\ C_6H_5OH,\ b \ mol\ CH_3OH\\
\Rightarrow \begin{cases}94a + 32b = 25,2\\\dfrac12(a+b) = \dfrac{4,48}{22,4}\end{cases}
\Rightarrow a = b= 0,2\ \rm mol\\
\Rightarrow \%m_{C_6H_5OH} = \dfrac{0,2\times 94\times 100\%}{25,2} = 74,6\%\\
2)\\
\rm X:\quad RCOOH\\
\rm RCOOH + Na \longrightarrow RCOONa + \dfrac12H_2\uparrow\\
0,2\rm mol \qquad \qquad \qquad \qquad \qquad 0,1\ \rm mol\\
\quad M_{axit} = \dfrac{12}{0,2}\\
\Leftrightarrow R + 45 = 60\\
\Leftrightarrow R = 15\quad \rm (CH_3)\\
\text{Vậy}\rm \ X:\quad CH_3COOH\\
3)\\
\rm Y:\quad R(COOH)_x\\
\rm R(COOH)_x + xNa \longrightarrow R(COONa)_x +\dfrac x2H_2\uparrow\\
\dfrac{0,2}{x}\ \rm mol\qquad \qquad \qquad \qquad \qquad\qquad 0,1\ mol\\
\quad M_{axit} = \dfrac{9}{\dfrac{0,2}{x}}\\
\Leftrightarrow R + 45x = 45x\\
\Leftrightarrow R = 0\\
\text{Vậy}\ \rm Y:\quad HOOC-COOH\\
4)\\
\rm A:\quad RCOOH\\
\rm RCOOH + Na \longrightarrow RCOONa + \dfrac12H_2\uparrow\\
0,1\rm mol \qquad \qquad \qquad \qquad \qquad 0,05\ \rm mol\\
\quad M_{axit} = \dfrac{7,2}{0,1}\\
\Leftrightarrow R + 45 = 72\\
\Leftrightarrow R = 27\ \rm (C_2H_3)\\
\text{Vậy}\ \rm A:\quad CH_2=CHCOOH
\end{array}\)