Đáp án:
\(\begin{array}{l}
C{\% _{NaCl}} = 0,35\% \\
C{\% _{HC{l_d}}} = 1,24\%
\end{array}\)
Giải thích các bước giải:
tác dụng vào 500g $HCl$
\(\begin{array}{l}
2Na + 2HCl \to 2NaCl + {H_2}\\
{n_{Na}} = \dfrac{{0,69}}{{23}} = 0,03mol\\
{m_{HCl}} = \dfrac{{500 \times 1,46}}{{100}} = 7,3g\\
{n_{HCl}} = \dfrac{{7,3}}{{36,5}} = 0,2mol\\
\dfrac{{0,03}}{2} < \dfrac{{0,2}}{2} \Rightarrow HCl\text{ dư}\\
{n_{{H_2}}} = \dfrac{{{n_{Na}}}}{2} = 0,015mol\\
{n_{NaCl}} = {n_{Na}} = 0,03mol\\
{m_{NaCl}} = 0,03 \times 58,5 = 1,755g\\
{n_{HCl{\rm{d}}}} = {n_{HCl}} - {n_{Na}} = 0,17mol\\
{m_{HC{l_d}}} = 0,17 \times 36,5 = 6,205g\\
{m_{{\rm{dd}}spu}} = 0,69 + 500 - 0,03 = 500,66g\\
C{\% _{NaCl}} = \dfrac{{1,755}}{{500,66}} \times 100\% = 0,35\% \\
C{\% _{HC{l_d}}} = \dfrac{{6,205}}{{500,66}} \times 100\% = 1,24\%
\end{array}\)
