Question

Bài 2 ạ 

Answer 1

a) $n_{NaOH}  = \dfrac{10}{40} = 0,25(mol)$
$NaOH + HCl \to NaCl  + H_2O$

Ta thấy : 

$n_{NaOH} : 1 < n_{HCl} : 1$ nên $HCl$ dư

$n_{NaCl} = n_{HCl\ pư} =n_{NaOH} = 0,25(mol)$
$m_{NaCl} = 0,25.58,5 = 14,625(gam)$

$m_{NaOH\ dư} = (2 - 0,25).36,5 = 63,875(gam)$

b)

$C_{M_{NaCl}} = \dfrac{0,25}{0,5} = 0,5M$

$C_{M_{NaOH\ dư}} = \dfrac{2-0,25}{0,5}  = 3,5M$