@$\text{Ly}$

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

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

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

Theo PT, $n_{Na}$ $=$ $2.n_{H_2}$ $=$ $0,1$ $mol$

Và $n_{NaOH}$ $=$ $n_{Na}$ $=$ $0,1$ $mol$

$m_{Na}$ $=$ $0,1.23=2,3$ $(g)$

$→$ $m_{Na_2O}$ $=$ $14,7-2,3=12,4$ $(g)$

$n_{Na_2O}$ $=$ $\dfrac{12,4}{62}$ $=$ $0,2$ $mol$

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

Theo PT trên, $n_{NaOH}$ $=$ $2.n_{Na_2O}$ $=$ $0,4$ $mol$

$m_{NaOH}$ $=$ $40.(0,1+0,4)=20$ $(g)$

$m_{ddNaOH}$ $=$ $14,7+185,4-(0,05.2)=200$ $(g)$

`C%_(NaOH` $=$ $\dfrac{20}{200}$`.100%` $=$ $10$`%`

$nH_2$ = $\frac{1,12}{22,4}$ = $0,05$ $mol$
$2Na$ + $2H_2O$ → $2NaOH$ + $H_2$

$0,1$                           $0,1$           $0,05$

→ $nNa$ = $0,1$ mol

Có $mNa$ + $mNa_2O$ = $14,7$

⇒ $0,1$ × $23$ + $mNa_2O$ = $14,7$
⇔ $mNa_2O$  = $12,4g$
⇒ $nNa_2O$ = $\frac{12,4}{62}$  = $0,2$ $mol$

$Na_2O$ + $H_2O$ → $2NaOH$

$0,2$                                $0,4$

⇒ ∑$nNaOH$ = $0,5$ mol

Có $mddspu$ = $mhh$ + $mH_2O$ - $mH_2$

                        =  $14,7$ + $185,4$ - $0,05$ × $2$

                        = $200g$

$C$% = $\frac{mNaOH}{mddspu}$  × $100$% =  $\frac{0,5 . 40}{200}$ × $100$% = $10$%