Đáp án:

\(\begin{array}{l}
16\,A\\
17\,B\\
18D\\
19\,C\\
20\,A\\
21A\\
22\,B\\
23\,B
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
16)\\
CTHH:xN{a_2}O.yCaO.zSi{O_2}\\
x:y:z = \dfrac{{13}}{{62}}:\dfrac{{11,72}}{{56}}:\dfrac{{75,28}}{{60}} = 0,21:0,21:1,25\\
 \Rightarrow x:y:z = 1:1:6\\
 \Rightarrow CTHH:N{a_2}O.CaO.6Si{O_2}\\
17)\\
Si{O_2} + {H_2}O \to {H_2}Si{O_3}\\
18)\\
2NaOH + Si{O_2} \to N{a_2}Si{O_3} + {H_2}O\\
{n_{Si{O_2}}} = \dfrac{{120}}{{60}} = 2\,mol\\
{n_{NaOH}} = 2{n_{Si{O_2}}} = 4\,mol\\
{m_{NaOH}} = 4 \times 40 = 160g\\
23)\\
2R + 2{H_2}O \to 2ROH + {H_2}\\
{n_{{H_2}}} = \dfrac{{1,12}}{{22,4}} = 0,05\,mol \Rightarrow {n_R} = 0,05 \times 2 = 0,1\,mol\\
{M_R} = \dfrac{{3,1}}{{0,1}} = 31g/mol \Rightarrow hh:Na,K
\end{array}\)

Sasori

`\begin{array}{l} 16)\\ CTHH:xN{a_2}O.yCaO.zSi{O_2}\\ x:y:z = \dfrac{{13}}{{62}}:\dfrac{{11,72}}{{56}}:\dfrac{{75,28}}{{60}} = 0,21:0,21:1,25\\  \Rightarrow x:y:z = 1:1:6\\  \Rightarrow CTHH:N{a_2}O.CaO.6Si{O_2}\\ 17)\\ Si{O_2} + {H_2}O \to {H_2}Si{O_3}\\ 18)\\ 2NaOH + Si{O_2} \to N{a_2}Si{O_3} + {H_2}O\\ {n_{Si{O_2}}} = \dfrac{{120}}{{60}} = 2\,mol\\ {n_{NaOH}} = 2{n_{Si{O_2}}} = 4\,mol\\ {m_{NaOH}} = 4 \times 40 = 160g\\ 23)\\ 2R + 2{H_2}O \to 2ROH + {H_2}\\ {n_{{H_2}}} = \dfrac{{1,12}}{{22,4}} = 0,05\,mol \Rightarrow {n_R} = 0,05 \times 2 = 0,1\,mol\\ {M_R} = \dfrac{{3,1}}{{0,1}} = 31g/mol \Rightarrow hh:Na,K \end{array}`