$D$
$2Na+2H_2O\to 2NaOH+H_2$
$Al_2(SO_4)_3+6NaOH\to 2Al(OH)_3\downarrow+3Na_2SO_4$
$Al(OH)_3+NaOH\to NaAlO_2+2H_2O$
(do $NaOH$ dư nên hòa tan hết kết tủa trắng $Al(OH)_3$)
D
2Na+2H2O->2NaOH+H2
6NaOH+Al2(SO4)3->2Na2SO4+3Al(OH)3
AL(OH)3+NaOH->NaAlO2+2H2O