Đáp án:

 60g

40%

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
nNa = \dfrac{m}{M} = \dfrac{{34,5}}{{23}} = 1,5\,mol\\
n{H_2}O = \dfrac{m}{M} = \dfrac{{117}}{{18}} = 6,5\,mol\\
2Na + 2{H_2}O \to 2NaOH + {H_2}\\
\dfrac{{1,5}}{2} < \dfrac{{6,5}}{2} \Rightarrow \text{ tính theo Na}\\
nNaOH = nNa = 1,5\,mol\\
mNaOH = n \times M = 1,5 \times 40 = 60g\\
b)\\
m{\rm{dd}}spu = 34,5 + 117 - 0,75 \times 2 = 150g\\
C\% NaOH = \dfrac{{mct}}{{m{\rm{dd}}}} \times 100\%  = \dfrac{{60}}{{150}} \times 100\%  = 40\% 
\end{array}\)

$n_{Na}=34,5/23=1,5mol$

$PTHH :$

$2Na+2H_2O\to 2NaOH+H_2↑$

$\text{a/Theo pt :}$

$n_{NaOH}=n_{Na}=1,5mol$

$⇒m_{NaOH}=1,5.40=60g$

$\text{b/Theo pt :}$

$n_{H_2}=1/2.n_{Na}=1/2.1,5=0,75mol$

$C\%_{NaOH}=\dfrac{60}{34,5+117-0,75.2}=40\%$