Gia sử : V = 22.4 (l)
nH2 = 1 mol
Đặt : nNa = x mol
nAl = y mol
2Na + 2H2O --> 2NaOH + H2
x________________x_____0.5x
NaOH + Al + H2O --> NaAlO2 + 3/2H2
x____________________________1.5x
nH2 = 0.5x + 1.5x = 1 (1)
<=> x = 0.5
2Na + 2H2O --> 2NaOH + H2
x________________x_____0.5x
NaOH + Al + H2O --> NaAlO2 + 3/2H2
________y____________________1.5y
<=> 0.5x + 1.5y = 1.75
=> y = 1
mNa = 0.5*23=11.5 g
mAl = 27 g
%Na = 29.09%
%Al = 70.91%