Mixell, your reaction is the same as blogfast25's. You see, in your reaction you get the same amount of hydrogen as in blogfast25's reaction for the
same amount of aluminium and the same amount of NaOH is used. The only difference is the amount of water taken up.
The compound NaAl(OH)4 can be written as NaAlO2.2H2O. The real compound probably is none of these exactly, it can best be described as NaAlO2.xH2O,
with x some indefinite value. |