Vì hh tan hết => NaOH dư, Al tan hết
Gọi \(\left\{{}\begin{matrix}n_{Na}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\)
=> 23a + 27b = 10 (1)
PTHH:
\(2Na+2H_2O\rightarrow2NaOH+H_2\)
a------------------->a----------->0,5a
\(2NaOH+2Al+2H_2O\rightarrow2NaAlO_2+3H_2\)
b<----------b------------------->b----------->1,5b
=> 0,5a + 1,5b = 0,4 (2)
(1),(2) => a = b = 0,2
\(n_{NaOH\left(dư\right)}=0,2-0,2=0\left(mol\right)\)
=> NaOH pư hết
ddX gồm: \(NaAlO_2:0,2\left(mol\right)\)
PTHH:
\(NaAlO_2+HCl+H_2O\rightarrow Al\left(OH\right)_3+NaCl\)
0,2-------->0,2
\(V_{dd.HCl}=\dfrac{0,2}{1}=0,2\left(l\right)\)