Đáp án:

`a)` $\begin{cases} m_{NaOH}=32(g)\\m_{H_2}=0,8(g) \end{cases}$

`b)` $m_{\text{dd spứ}}$`=149,2(g)`

`c)` `C%`$_{\text{dd NaOH}}$`~~21,45%`

Giải thích các bước giải:

`Pt : 2Na + 2H_2O -> 2NaOH + H_2\uparrow`

`a)`

`n_{Na}={18,4}/23=0,8(mol)`

Theo pt : $\begin{cases} n_{NaOH}=n_{Na}=0,8(mol)\\n_{H_2}=\dfrac{1}{2}.n_{Na}=0,4(mol) \end{cases}$

`=>` $\begin{cases} m_{NaOH}=0,8.40=32(g)\\m_{H_2}=0,4.2=0,8(g) \end{cases}$

`b)`

`m_{H_2O}=131,6.1=131,6(g)`

`=>` $m_{\text{dd spứ}}$`=18,4+131,6-0,8=149,2(g)` 

`c)`

`C%`$_{\text{dd NaOH}}$ `={32.100%}/{149,2}~~21,45%`

`#Ly`

$a)$ 

$n_{Na}$ $=$ $\dfrac{m}{M}$ $=$ $\dfrac{18,4}{23}$ $=$ $0,8$ $(mol)$

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

Theo PT, $n_{NaOH}$ $=$ $n_{Na}$ $=$ $0,8$ $(mol)$

$m_{NaOH}$ $=$ `n.M=0,8.40=32` $(g)$

Theo PT, $n_{H_2}$ $=$ $\dfrac{1}{2}n_{Na}$ $=$ $0,4$ $(mol)$

$m_{H_2}$ $=$ $n.M=0,4.2=0,8$ $(g)$

$b)$

Do khối lượng riêng nước luôn bằng $1$

$⇒$ $m_{H_2O}$ $=$ $131,6$ $(g)$

$m_{ddNaOH}=m_{Na}+m_{H_2O}-m_{H_2}=18,4+131,6-0,8=149,2$ $(g)$

$c)$

`C%_(NaOH` $=$ `(m_{NaOH})/(m_{ddNaOH}).100%=(32)/(149,2).100%≈21,45%`