Giải thích các bước giải:

\(\begin{array}{l}
2.\\
{n_{C{O_2}}} = 0,05mol\\
{n_{KOH}} = 0,125mol\\
 \to \frac{{{n_{KOH}}}}{{{n_{C{O_2}}}}} = \frac{{0,125}}{{0,05}} = 2,5\\
C{O_2} + 2KOH \to {K_2}C{O_3} + {H_2}O\\
CaC{l_2} + {K_2}C{O_3} \to CaC{O_3} + 2KCl\\
{n_{{K_2}C{O_3}}} = {n_{C{O_2}}} = 0,05mol\\
 \to {n_{CaC{O_3}}} = {n_{{K_2}C{O_3}}} = 0,05mol\\
 \to {m_{CaC{O_3}}} = 5g\\
6.\\
{n_{Ca{{(OH)}_2}}} = 0,1mol\\
{n_{CaC{O_3}}} = 0,002mol\\
 \to {n_{Ca{{(OH)}_2}}} > {n_{CaC{O_3}}}
\end{array}\)

=> Tạo 2 muối

\(\begin{array}{l}
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O(1)\\
2C{O_2} + Ca{(OH)_2} \to Ca{(HC{O_3})_2}(2)\\
{n_{Ca{{(OH)}_2}(1)}} = {n_{CaC{O_3}}} = 0,002mol\\
 \to {n_{Ca{{(OH)}_2}(2)}} = 0,008mol\\
 \to {n_{C{O_2}}} = {n_{Ca{{(OH)}_2}(1)}} + 2{n_{Ca{{(OH)}_2}(2)}} = 0,018mol\\
 \to {V_{C{O_2}}} = 0,4032l
\end{array}\)

\(\begin{array}{l}
3.\\
{n_{C{O_2}}} = 0,04mol\\
{n_{Ca{{(OH)}_2}}} = 0,03mol\\
 \to \dfrac{{{n_{Ca{{(OH)}_2}}}}}{{{n_{C{O_2}}}}} = \dfrac{{0,03}}{{0,04}} = 0,75
\end{array}\)

=> Tạo 1 muối: \(CaC{O_3}\)

\(\begin{array}{l}
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O\\
{n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,04mol\\
 \to {m_{CaC{O_3}}} = 4g
\end{array}\)