Bài 1 :
$Na_2O + H_2O \to NaOH \\ 2KOH + CO_2 \to K_2CO_3 + H_2O\\ Cu(OH)_2 + H_2SO_4 \to CuSO_4 + 2H_2O \\ 2KOH + CuCl_2 \to Cu(OH)_2 + 2KCl \\ Ba(OH)_2 + H_2SO_4 \to BaSO_4 + 2H_2O \\ 2Al(OH)_3 \xrightarrow{t^o} Al_2O_3 + 3H_2O \\ CO_2 + Al(OH)_3 \to \text{Không phản ứng} $
Bài 2 :
a)
$n_{Na_2O} = \dfrac{15,5}{62} = 0,25(mol)$
$Na_2O + H_2O \to 2NaOH$
$n_{NaOH} = 2n_{Na_2O} = 0,25.2 = 0,5(mol)$
$C_{M_{NaOH}} = \dfrac{0,5}{0,5} = 1M$
b)
$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
$n_{H_2SO_4\ pư} = 0,5n_{NaOH} = 0,25(mol)$
$m_{dd\ H_2SO_4} = \dfrac{0,25.98}{20\%} = 122,5(gam)$
$V_{dd\ H_2SO_4} = \dfrac{122,5}{1,14} = 107,46(ml)$
