Đáp án  + Giải thích các bước giải:

 PTHH:
$2Na + 2H_2O \longrightarrow 2NaOH + H_2$

$Na_2O + H_2O \longrightarrow 2NaOH$

$Ba + 2H_2O \longrightarrow Ba(OH)_2 + H_2$

$BaO + H_2O \longrightarrow Ba(OH)_2$

$n_{Ba(OH)_2} = \dfrac{m}{M} = \dfrac{20,52}{137 + 2 . (16 + 1)} = \dfrac{20,52}{171} = 0,12\text{ mol}$

$n_{H_2} = \dfrac{1,12}{22,4} = 0,05\text{ mol}$

Gọi $n_{NaOH} = a\text{ mol}$

Ta có:
$n_H = 2n_{Ba(OH)_2} + n_{NaOH} + 2n_{H_2} = 2 . 0,12 + 2 . 0,05 + a = 0,34 + a$

$\Rightarrow n_{H_2O} = \dfrac{0,34 + a}{2} = 0,17 + 0,5a$

Bảo toàn khối lượng, ta có:
$21,9 + 18(0,17 + 0,5a) = 20,52 + 40a + 0,05 . 2$

$\Rightarrow 21,9 + 3,06 + 9a = 20,52 + 40a + 0,1$

$\Rightarrow 24,96 + 9a = 20,62 + 40a$

$\Rightarrow 24,96 - 20,62 = 40a - 9a$

$\Rightarrow 4,34 = 31a$

$\Rightarrow a = 0,14\text{ mol}$

Đáp án:

$0,14(mol)$

Giải thích các bước giải:

$2Na+2H_2O\to 2NaoH+H_2\\ Ba+2H_2O\to Ba(OH)_2+H_2\\ Na_2O+H_2O\to 2NaOH\\ BaO+H_2O\to Ba(OH)_2$

$n_{Ba(OH)_2}=\dfrac{20,52}{22,4}=0,12(mol)\\ n_{H_2}=\dfrac{1,12}{22,4}=0,05(mol)\\ \text{Quy A thành}\ Na;\ Ba;\ O\\ \xrightarrow{\text{BTNT Ba}}\ n_{Ba}=n_{Ba(OH)_2}=0,12(mol)\\ \Rightarrow 23x+16y+0,12.137=21,9\\ \Rightarrow 23x+16y=5,46\ (1)\\ \xrightarrow{\text{BT e}}\ n_{Na}+2.n_{Ba}=2.n_O+2.n_{H_2}\\ \Rightarrow x+0,24=2y+0,1\\ \Rightarrow x-2y=-0,14\ (2)\\ \xrightarrow{\text{Từ (1);(2}}\begin{cases}x=0,14\\y=0,14\end{cases}\\ \xrightarrow{\text{BTNT Na}}\ n_{NaOH}=n_{Na}=0,14(mol)\\ \Rightarrow n_{NaOH}=0,14$