Gọi $n_{Na} = a(mol) ; n_{Al} = b(mol) \Rightarrow 23a + 27b = 7,3(1)$
\(2Na+2H_2O\text{→}2NaOH+H_2\)
a 0,5a (mol)
\(2Al+2NaOH+2H_2O\text{→}2NaAlO_2+3H_2\)
b 1,5b (mol)
Ta có :
$7,3 + 50 = 56,8 + (0,5a + 1,5b).2(2)$
Từ (1)(2) suy ra a = 0,2 ; b = 0,1
$m_{Al} = 0,1.27 = 2,7(gam)$