Đáp án:
\(\begin{array}{l}
{\rm{a}})C{M_{HCl}} = 1M\\
b)\\
\% {m_{N{a_2}C{O_3}}} = 42,4\% \\
\% {m_{KCl}} = 57,6\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
N{a_2}C{O_3} + 2HCl \to 2NaCl + C{O_2} + {H_2}O\\
{\rm{a}})\\
{n_{C{O_2}}} = 0,01mol\\
\to {n_{HCl}} = 2{n_{C{O_2}}} = 0,02mol\\
\to C{M_{HCl}} = \dfrac{{0,02}}{{0,02}} = 1M\\
b)\\
{n_{N{a_2}C{O_3}}} = {n_{C{O_2}}} = 0,01mol \to {m_{N{a_2}C{O_3}}} = 1,06g\\
\to \% {m_{N{a_2}C{O_3}}} = \dfrac{{1,06}}{{2,5}} \times 100\% = 42,4\% \\
\to \% {m_{KCl}} = 100\% - 42,4\% = 57,6\%
\end{array}\)