a,
$2C_2H_5OH+2Na\to 2C_2H_5ONa+H_2$
$2Na+2H_2O\to 2NaOH+H_2$
b,
$V_{C_2H_5OH}=10.96\%= 9,6ml$
$\Rightarrow m_{C_2H_5OH}=9,6.0,8=7,68g$
c,
$n_{C_2H_5OH}=\frac{7,68}{46}= 0,167 mol$
$V_{H_2O}=10-9,6=0,4ml$
$\Rightarrow m_{H_2O}=0,4g$
$n_{H_2O}=\frac{0,4}{18}=0,022 mol$
$n_{H_2}= 0,5n_{C_2H_5OH}+0,5n_{H_2O}= 0,0945mol$
$\Rightarrow V_{H_2}=0,0945.22,4=2,1168l$
