Question

Answer 1

$V_{C_2H_5OH} = 60.\dfrac{40}{100} = 24(ml)$

$V_{H_2O} = 60 - 24 = 36(ml)$

$m_{C_2H_5OH} = D.V = 0,8.24 = 19,2(gam) ; m_{H_2O} = 36.1 = 36(gam)$

Suy ra : $n_{C_2H_5OH} = \dfrac{48}{115}(mol);  n_{H_2O} = 2(mol)$
$C_2H_5OH + Na \to C_2H_5ONa + \dfrac{1}{2}H_2$
$H_2O + Na \to NaOH + \dfrac{1}{2}H_2$

$n_{H_2} = \dfrac{1}{2}n_{C_2H_5OH} + \dfrac{1}{2}n_{H_2O} = \dfrac{139}{115}(mol)$
$V_{H_2} = \dfrac{139}{115}.22,4 = 27,07(lít)$