a)
$Ca + 2H_2O \to Ca(OH)_2 + H_2$
$Ca(OH)_2 + Na_2CO_3 \to CaCO_3 + 2NaOH$
b)
$Ba + 2H_2O \to Ba(OH)_2 + H_2$
$2NaHSO_4 + Ba(OH)_2 \to BaSO_4 + Na_2SO_4 + 2H_2O$
c)
$2Na + 2H_2O \to 2NaOH + H_2$
$AlCl_3 + 3NaOH \to Al(OH)_3 + 3NaCl$
Có thể xảy ra :
$Al(OH)_3 + NaOH \to NaAlO_2 + 2H_2O$
\(a.\)
\(Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\)
\(Ca\left(OH\right)_2+Na_2CO_3\rightarrow CaCO_3+2NaOH\)
\(b.\)
\(Ba+2H_2O\rightarrow Ba\left(OH\right)_2+H_2\)
\(Ba\left(OH\right)_2+2NaHSO_4\rightarrow BaSO_4+Na_2SO_4+H_2O\)
\(c.\)
\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)
\(AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3+NaCl\)
\(NaOH+Al\left(OH\right)_3\rightarrow NaAlO_2+2H_2O\)
Tất cả pư đều xảy ra. Nhưng vì là KL kiềm nên trc khi pư với dd muối thì chúng pư vs nước trong dd trước
a) Ca+2H2O-->Ca(OH)2+2H2
Ca(OH)2+Na2CO3-->CaCO3+2NaOH
b) Ba+2H2O-->Ba(OH)2+2H2
Ba(OH)2+2NaHSO4-->BaSO4+Na2SO4+2H2O
c) 2Na+2H2O-->2NaOH+H2
3NaOH+AlCl3--->Al(OH)3+3NaCl
