Đáp án:
\(\begin{array}{l}
a)\\
{m_{NaOH}} = 40g\\
b)\\
C{\% _{NaOH}} = 68,97\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
N{a_2}O + {H_2}O \to 2NaOH\\
{n_{N{a_2}O}} = \dfrac{m}{M} = \dfrac{{31}}{{62}} = 0,5mol\\
{n_{NaOH}} = 2{n_{N{a_2}O}} = 1mol\\
{m_{NaOH}} = n \times M = 1 \times 40 = 40g\\
b)\\
{m_{dd}} = {m_{N{a_2}O}} + {m_{{H_2}O}} = 31 + 27 = 58g\\
C{\% _{NaOH}} = \frac{{40}}{{58}} \times 100\% = 68,97\%
\end{array}\)