Đáp án:
\(\begin{array}{l}
a.\\
\% {m_{Na}} = 26,34\% \\
\% {m_{N{a_2}O}} = 73,66\% \\
b.\\
C{\% _{NaOH}} = 8,83\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
Na + {H_2}O \to NaOH + \frac{1}{2}{H_2}\\
N{a_2}O + {H_2}O \to 2NaOH\\
{n_{{H_2}}} = 0,15mol\\
\to {n_{Na}} = 2{n_{{H_2}}} = 0,15mol\\
\to {m_{Na}} = 3,45g\\
\to {m_{N{a_2}O}} = 9,65g \to {n_{N{a_2}O}} = 0,16mol
\end{array}\)
\(\begin{array}{l}
a.\\
\% {m_{Na}} = \dfrac{{3,45}}{{13,1}} \times 100\% = 26,34\% \\
\% {m_{N{a_2}O}} = 100\% - 26,34\% = 73,66\%
\end{array}\)
\(\begin{array}{l}
b.\\
{n_{NaOH}} = {n_{Na}} + 2{n_{N{a_2}O}} = 0,47mol\\
\to {m_{NaOH}} = 18,8g\\
\to {m_{{\rm{dd}}}} = {m_{hỗnhợp}} + {m_{{H_2}O}} - {m_{{H_2}}} = 212,8g\\
\to C{\% _{NaOH}} = \dfrac{{18,8}}{{212,8}} \times 100\% = 8,83\%
\end{array}\)
