Đáp án + Giải thích các bước giải:

 a) PTHH:
$2Na + 2H_2O \longrightarrow 2NaOH + H_2 \uparrow$
$Na_2O + H_2O \longrightarrow 2NaOH$

$n_{H_2} = \dfrac{V}{22,4} = \dfrac{2,24}{22,4} = 0,1\text{ mol}$
Theo PTHH:
$n_{Na} = 2n_{H_2} = 2 . 0,1 = 0,2\text{ mol}$
$\Rightarrow m_{Na} = n . M = 0,2 . 23 = 4,6g$

$m_{Na_2O} = 12,4 - 4,6 = 7,8g$

b) $\%_{Na} = \dfrac{4,6}{12,4} . 100\% \approx 37,1\%$

$\%_{Na_2O} =100 - 37,1 = 62,9\%$
c) $n_{Na_2O} = \dfrac{m}{M} = \dfrac{7,8}{2 . 23 + 16} = \dfrac{7,8}{62} \approx 0,126\text{ mol}$

Theo PTHH:
$n_{NaOH} = n_{Na} + 2n_{Na_2O} = 0,2 + 2 . 0,126 = 0,452\text{ mol}$

$\Rightarrow m_{NaOH} =n . M = 0,452 . (23 + 16 +1) = 0,452 . 40 = 18,08g$

@$\text{Ly}$

$n_{H_2}$ $=$ $\dfrac{2,24}{22,4}$ $=$ $0,1$ $mol$

Do trong hỗn hợp chỉ có $Na$ tác dụng với $H_2O$ tạo khí, nên ta có:

PTHH: $2Na$ $+$ $2H_2O$ `->` $2NaOH$ $+$ $H_2↑$

Theo PT, $n_{Na}$ $=$ $2.n_{H_2}$ $=$ $0,2$ $mol$

Và $n_{Na}$ $=$ $n_{NaOH}$ $=$ $0,2$ $mol$

$a)$ $m_{Na}$ $=$ $0,2.23=4,6$ $(g)$

$m_{Na_2O}$ $=$ $12,4-4,6=7,8$ $(g)$

$b)$ `%m_{Na` $=$ $\dfrac{4,6}{12,4}$`.100%` $=$ $37,09$`%`

`%m_{Na_2O` $=$ $100-37,09=62,91$`%`

$n_{Na_2O}$ $=$ $\dfrac{7,8}{62}$ $=$ $0,125$ $mol$

PTHH: $Na_2O$ $+$ $2H_2O$ `->` $2NaOH$ $+$ $H_2O$

Theo PT, $n_{NaOH}$ $=$ $2.n_{Na_2O}$ $=$ $0,25$ $mol$

$c)$ $m_{NaOH}$ $=$ $40.(0,2+0,25)=18$ $(g)$