Đáp án:
\(\begin{array}{l}
b){V_{C{O_2}}} = 1,12l\\
c)C{\% _{{\rm{dd}}HCl}} = 3,65\% \\
d)\\
{m_{CaC{O_3}}} = 5g\\
C{M_{Ca{{(OH)}_2}}} = 0,25M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a){K_2}C{O_3} + 2HCl \to 2KCl + C{O_2} + {H_2}O\\
b)\\
{n_{{K_2}C{O_3}}} = 0,05mol\\
\to {n_{C{O_2}}} = {n_{{K_2}C{O_3}}} = 0,05mol\\
\to {V_{C{O_2}}} = 1,12l\\
c)\\
{n_{HCl}} = 2{n_{{K_2}C{O_3}}} = 0,1mol\\
\to {m_{HCl}} = 3,65g\\
\to C{\% _{{\rm{dd}}HCl}} = \dfrac{{3,65}}{{100}} \times 100\% = 3,65\% \\
d)\\
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O\\
{n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,05mol\\
\to {m_{CaC{O_3}}} = 5g\\
{n_{Ca{{(OH)}_2}}} = {n_{C{O_2}}} = 0,05mol\\
\to C{M_{Ca{{(OH)}_2}}} = \dfrac{{0,05}}{{0,2}} = 0,25M
\end{array}\)
