Đáp án:
a) $\% {m_{N{a_2}C{O_3}}} = 54,25\% ;\% {m_{KCl}} = 45,75\% $
b) 100 g
Giải thích các bước giải:
a) $N{a_2}C{O_3} + 2HCl \to 2NaCl + C{O_2} + {H_2}O$
$\begin{gathered}
{n_{C{O_2}}} = \dfrac{{5,6}}{{22,4}} = 0,25mol \Rightarrow {n_{N{a_2}C{O_3}}} = 0,25mol \hfill \\
\Rightarrow {m_{N{a_2}C{O_3}}} = 0,25.106 = 26,5g \hfill \\
\Rightarrow \% {m_{N{a_2}C{O_3}}} = \dfrac{{26,5}}{{48,85}}.100\% = 54,25\% \hfill \\
\Rightarrow \% {m_{KCl}} = 100 - 54,25 = 45,75\% \hfill \\
\end{gathered} $
b)
Theo PTHH: ${n_{HCl}} = 2{n_{C{O_2}}} = 2.0,25 = 0,5mol$
$ \Rightarrow {m_{ddHCl}} = \dfrac{{0,5.36,5}}{{18,25\% }} = 100g$
