Đáp án:
a) C% = 71,875%

Giải thích các bước giải:

a) 12,8g X$\left \{ {{C2H5OH} \atop {H2O}} \right.$  pư vs K dư
K + H2O$\longrightarrow$ KOH + $\frac{1}{2}$ H2$\uparrow$

C2H5OH + K $\longrightarrow$ C2H5OK + $\frac{1}{2}$ H2 $\uparrow$
nH2 = 4,48/22,4 =0,2(mol)
$\longrightarrow$ nX = 2.nH2 =2.0,2=0,4 mol
Đặt $\left \{ {{nC2H5OH=x(mol)} \atop {nH2O =y(mol)}} \right.$ (x,y>0)
$\longrightarrow$ nX = x+y = 0,4 (mol)         
mX = mC2H5OH + mH2O = 46x + 18y = 12,8(g)
$\longrightarrow$ $\left \{ {{x=0,2(mol)} \atop {y=0,2(mol)}} \right.$ 
$\longrightarrow$ C%C2H5OH = $\frac{mC2H5OH}{mX}$ .100%
= $\frac{0,2.46}{12,8}$ .100%=71,875%

b)  nH2= 0,2(mol)

Na + H2O$\longrightarrow$ NaOH + $\frac{1}{2}$ H2$\uparrow$

C2H5OH + Na $\longrightarrow$ C2H5ONa + $\frac{1}{2}$ H2 $\uparrow$
--> nX = 0,4 mol