Đáp án:
$\begin{gathered}
C{\% _{NaCl}} = 2,76\% \hfill \\
C{\% _{HCldu}} = 3\% \hfill \\
\end{gathered} $
Giải thích các bước giải:
${n_{N{a_2}C{O_3}.10{H_2}O}} = \dfrac{{14,3}}{{286}} = 0,05mol$
${n_{HCl}} = \dfrac{{200}}{{36,5}}.\dfrac{5}{{100}} = 0,274mol$
$N{a_2}C{O_3} + 2HCl \to 2NaCl + C{O_2} + {H_2}O$
Theo PTHH: ${n_{HClpu}} = 2{n_{N{a_2}C{O_3}}} = 0,05.2 = 0,1mol$
${n_{NaCl}} = 2{n_{N{a_2}C{O_3}}} = 0,1mol;{n_{C{O_2}}} = {n_{N{a_2}C{O_3}}} = 0,05mol$
${n_{HCldu}} = 0,274 - 0,1 = 0,174mol$
Bảo toàn khối lượng:
$\begin{gathered}
{m_{tt}} + {m_{ddHCl}} = {m_{ddsau}} + {m_{C{O_2}}} \hfill \\
\Rightarrow {m_{ddsau}} = 14,3 + 200 - 0,05.44 = 212,1g \hfill \\
\end{gathered} $
$\begin{gathered}
C{\% _{NaCl}} = \dfrac{{0,1.58,5}}{{212,1}}.100\% = 2,76\% \hfill \\
C{\% _{HCldu}} = \dfrac{{0,174.36,5}}{{212,1}}.100\% = 3\% \hfill \\
\end{gathered} $
