$a)\begin{cases}n_{H_2O}=0,345 mol \\ n_{CO_2}=0,345 mol \end{cases}\Rightarrow n_{H_2O}=n_{CO_2}\Rightarrow $este no đơn chức mạch hở !$b)m_{KOH}=7,7+4,025-6,825=4,9 gam\Rightarrow n_{KOH}=0,0875 mol$
$\Rightarrow M_{ruou}=\frac{4,025}{0,0875}=46\Rightarrow C_2H_5OH$
$\Rightarrow \overline{M}_{este}=\frac{6,825}{0,0875}=78\Rightarrow \begin{cases}HCOOC_2H_5:x mol \\ CH_3COOC_2H_5:y mol \end{cases} $
$\Rightarrow \begin{cases}74x+88y=6,825 \\ x+y=0,0875 \end{cases}\Rightarrow \begin{cases}x=0,0625 mol \\ y=0,025 mol \end{cases}\Rightarrow \begin{cases}m_{HCOOC_2H_5}=4,625 gam \\ m_{CH_3COOC_2H_5}=2,2 gam \end{cases}$