Em tham khảo nha:

\(\begin{array}{l}
a)\\
{n_{NaOH}} = V \times {C_M} = 0,1 \times 1 = 0,1\,mol\\
{n_{C{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{1,68}}{{22,4}} = 0,075\,mol\\
T = \dfrac{{{n_{NaOH}}}}{{{n_{C{O_2}}}}} = \dfrac{{0,1}}{{0,075}} = 1,333\\
1 < T < 2 \Rightarrow\text{ Tạo cả 2 muối} \\
2NaOH + C{O_2} \to N{a_2}C{O_3} + {H_2}O(1)\\
N{a_2}C{O_3} + C{O_2} + {H_2}O \to 2NaHC{O_3}(2)\\
{n_{C{O_2}(1)}} = {n_{N{a_2}C{O_3}(1)}} = \dfrac{{0,1}}{2} = 0,05\,mol\\
{n_{C{O_2}(2)}} = {n_{N{a_2}C{O_3}(2)}} = 0,075 - 0,05 = 0,025\,mol\\
{n_{N{a_2}C{O_3}}} = 0,05 - 0,025 = 0,025\,mol\\
{n_{NaHC{O_3}}} = 0,025 \times 2 = 0,05\,mol\\
m = {m_{N{a_2}C{O_3}}} + {m_{NaHC{O_3}}} = 0,025 \times 106 + 0,05 \times 84 = 6,85g\\
b)\\
{n_{NaOH}} = V \times {C_M} = 0,1 \times 1 = 0,1\,mol\\
{n_{C{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{1,12}}{{22,4}} = 0,05\,mol\\
T = \dfrac{{{n_{NaOH}}}}{{{n_{C{O_2}}}}} = \dfrac{{0,1}}{{0,05}} = 2\\
T = 2 \Rightarrow\text{ Tạo muối $Na_2CO_3$} \\
2NaOH + C{O_2} \to N{a_2}C{O_3} + {H_2}O\\
{n_{N{a_2}C{O_3}}} = {n_{C{O_2}}} = 0,05\,mol\\
 \Rightarrow {m_{N{a_2}C{O_3}}} = n \times M = 0,05 \times 106 = 5,3g\\
c)\\
{n_{NaOH}} = V \times {C_M} = 0,1 \times 1 = 0,1\,mol\\
{n_{C{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
T = \dfrac{{{n_{NaOH}}}}{{{n_{C{O_2}}}}} = \dfrac{{0,1}}{{0,15}} = 0,667\\
T < 1 \Rightarrow\text{ Tạo muối $NaHCO_3$} \\
NaOH + C{O_2} \to NaHC{O_3}\\
{n_{NaHC{O_3}}} = {n_{NaOH}} = 0,1\,mol\\
{m_{NaHC{O_3}}} = n \times M = 0,1 \times 84 = 8,4g
\end{array}\)