\(\begin{array}{l}
\rm C_2H_5OH \quad+\quad Na\quad \longrightarrow\quad C_2H_5ONa \quad+\quad \dfrac12H_2\uparrow\\
x\ \text{mol}\qquad \qquad \qquad \qquad \qquad \qquad\qquad\qquad\quad \dfrac12x\ \rm mol\\
\rm C_6H_5OH \quad+\quad Na\quad \longrightarrow\quad C_6H_5ONa \quad+\quad \dfrac12H_2\uparrow\\
y\ \text{mol}\qquad \qquad \qquad \qquad \qquad \qquad\qquad\qquad\quad \dfrac12y\ \rm mol\\
n_{H_2} = \dfrac{4,48}{22,4} = 0,2\ \rm mol\\
a)\\
hh\ ancol:\ x\ \text{mol}\ C_2H_5OH,\ y\ \text{mol}\ C_6H_5OH\\
\Rightarrow \begin{cases}46x + 94y = 16\\\dfrac12x + \dfrac12y = 0,2\end{cases}\Rightarrow \begin{cases}x = 0,45\\y = -0,05 \end{cases}\\
\Rightarrow \text{Sai số liệu}\\
b)\\
\rm C_2H_5OH + CuO \xrightarrow{\quad t^\circ\quad} CH_3CHO + Cu\downarrow + H_2O
\end{array}\)

a,

Gọi $x$, $y$ là số mol $C_2H_5OH$, $C_6H_5OH$

$\to 46x+94y=16$                         $(1)$

$n_{H_2}=\dfrac{4,48}{22,4}=0,2(mol)$

Ta có: $n_{OH}=2n_{H_2}$

$\to x+y=0,2.2=0,4$                       $(2)$

$(1)(2)\to x=0,45; y=-0,05$ (vô lí, không có đáp án)

b,

$C_2H_5OH+CuO\xrightarrow{{t^o}} CH_3CHO+Cu+H_2O$