Em tham khảo nha:
\(\begin{array}{l}
9)\\
\text{ Muối}\\
NaCl:Natri\,clorua\\
CaC{l_2}:Can\,xi\,clorua\\
\text{ Bazo}\\
Fe{(OH)_2}:\text{ Sắt II hidroxit}\\
NaOH:\,Natri\,hidroxit\\
10)\\
CaO + {H_2}O \to Ca{(OH)_2}\\
Ca{(OH)_2} + C{O_2} \to CaC{O_3} + {H_2}O\\
CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O\\
CaC{l_2} + 2AgN{O_3} \to 2AgCl + Ca{(N{O_3})_2}\\
11)\\
a)\\
CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O\\
{n_{C{O_2}}} = \dfrac{{0,448}}{{22,4}} = 0,02\,mol\\
{n_{HCl}} = 0,02 \times 2 = 0,04\,mol\\
{C_M}HCl = \dfrac{{0,04}}{{0,4}} = 0,1M\\
b)\\
{n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,02\,mol\\
\% {m_{CaC{O_3}}} = \dfrac{{0,02 \times 100}}{5} \times 100\% = 40\% \\
\% {m_{CaS{O_4}}} = 100 - 40 = 60\% \\
12)\\
{n_{C{O_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
{n_{NaOH}} = \dfrac{{200 \times 10\% }}{{40}} = 0,5\,mol\\
T = \dfrac{{{n_{NaOH}}}}{{{n_{C{O_2}}}}} = \dfrac{{0,5}}{{0,3}} = 1,67\\
1 < T < 1,67 \Rightarrow \text{ Tạo cả 2 muối}\\
2NaOH + C{O_2} \to N{a_2}C{O_3} + {H_2}O(1)\\
N{a_2}C{O_3} + C{O_2} + {H_2}O \to 2NaHC{O_3}(2)\\
{n_{N{a_2}C{O_3}}}(1) = {n_{C{O_2}}}(1) = \dfrac{{0,5}}{2} = 0,25\,mol\\
{n_{N{a_2}C{O_3}}}(2) = {n_{C{O_2}}}(2) = 0,3 - 0,25 = 0,05\,mol\\
{n_{N{a_2}C{O_3}}} = 0,25 - 0,05 = 0,2\,mol\\
{n_{NaHC{O_3}}} = 0,05 \times 2 = 0,1\,mol\\
{m_{N{a_2}C{O_3}}} = 0,2 \times 106 = 21,2g\\
{m_{NaHC{O_3}}} = 0,1 \times 84 = 8,4g
\end{array}\)
