1.D2. $V_{rươu} = 92.2/100 = 1,84$ lít => $m_{rượu} = 0,8.1,84.1000 = 1472$ gam, => $n_{rượu} = 32$ mol
$m_{H_2O(trong dd rượu)} = (2-1,84).1000 = 160 => n_{H_2O} = 8,89$ mol
$C_2H_5OH + Na => C_2H_5ONa + 1/2H_2$
$H_2O + Na => NaOH +1/2H_2$
$n_{H_2} = 32/2 + 8,89/2 = 20,44$ mol => $V = 457,96$ lít