$*)10 ml:\begin{cases}9,6 ml:C_2H_5OH \\ 0,4 ml: H_2O \end{cases}\Rightarrow \begin{cases}m_{C_2H_5OH}=7,68 gam \\ m_{H_2O}=0,4 gam \end{cases}$$\Rightarrow V_{H_2}=22,4\times \frac{1}{2}\times (\frac{7,68}{46}+\frac{0,4}{18})=2,1188\approx 2,12 lit\Rightarrow Chon:D$