Đáp án:

1, \(N{a_2}O,{K_2}O,CaO,CuO,A{l_2}{O_3},C{O_2},S{O_2},{P_2}{O_5}\)

a,

\(\begin{array}{l}
N{a_2}O + {H_2}O \to 2NaOH\\
{K_2}O + {H_2}O \to 2KOH\\
CaO + {H_2}O \to Ca{(OH)_2}\\
C{O_2} + {H_2}O \to {H_2}C{O_3}\\
S{O_2} + {H_2}O \to {H_2}S{O_3}\\
{P_2}{O_5} + 3{H_2}O \to 2{H_3}P{O_4}
\end{array}\)

b,

Hóa đỏ: \({H_2}C{O_3}\),\({H_2}S{O_3}\),\({H_3}P{O_4}\)

Hóa xanh:\(NaOH\),\(KOH\),\(Ca{(OH)_2}\)

2,

\(\begin{array}{l}
Na + {H_2}O \to NaOH + \dfrac{1}{2}{H_2}\\
K + {H_2}O \to KOH + \dfrac{1}{2}{H_2}\\
{n_{Na}} = 0,2mol\\
{n_K} = 0,1mol\\
{n_{{H_2}}} = \dfrac{1}{2} \times ({n_{Na}} + {n_K}) = 0,15mol \to V = 3,36l\\

\end{array}\)

c, Dung dịch sau phản ứng làm quỳ tím hóa xanh