Em tham khảo nha:

\(\begin{array}{l}
1)\\
a)\,{K_2}O,CaO,S{O_2}\\
{K_2}O + {H_2}O \to 2KOH\\
CaO + {H_2}O \to Ca{(OH)_2}\\
S{O_2} + {H_2}O \to {H_2}S{O_3}\\
b)\,{K_2}O,CaO,CuO,FeO\\
{K_2}O + 2HCl \to 2KCl + {H_2}O\\
CaO + 2HCl \to CaC{l_2} + {H_2}O\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
FeO + 2HCl \to FeC{l_2} + {H_2}O\\
c)\,S{O_2}\\
2NaOH + S{O_2} \to N{a_2}S{O_3} + {H_2}O\\
3)\\
a)\\
C{O_2} + 2KOH \to {K_2}C{O_3} + {H_2}O\\
b)\\
{n_{C{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
{n_{KOH}} = 2{n_{C{O_2}}} = 0,2\,mol\\
{C_M}(KOH) = \dfrac{n}{V} = \dfrac{{0,2}}{{0,1}} = 2M\\
c)\\
{n_{{K_2}C{O_3}}} = {n_{C{O_2}}} = 0,1\,mol\\
{m_{{K_2}C{O_3}}} = n \times M = 0,1 \times 138 = 13,8g
\end{array}\)

2)

Cho 2 chất trên vào dd HCl

- Tan và có khí thoát ra là $CaCO_3$

- Tan là CaO

\(\begin{array}{l}
CaO + 2HCl \to CaC{l_2} + {H_2}O\\
CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O
\end{array}\)