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