Em tham khảo nha :
\(\begin{array}{l}
C + {O_2} \to C{O_2}\\
{n_C} = \dfrac{{12}}{{12}} = 1mol\\
{n_{C{O_2}}} = {n_C} = 1mol\\
a)\\
NaOH + C{O_2} \to NaHC{O_3}\\
{n_{NaOH}} = {n_{C{O_2}}} = 1mol\\
{V_{NaOH}} = \dfrac{1}{2} = 0,5l\\
b)\\
2NaOH + C{O_2} \to N{a_2}C{O_3} + {H_2}O\\
{n_{NaOH}} = 2{n_{C{O_2}}} = 2mol\\
{V_{NaOH}} = \dfrac{2}{2} = 1l\\
c)\\
NaOH + C{O_2} \to NaHC{O_3}\\
2NaOH + C{O_2} \to N{a_2}C{O_3} + {H_2}O\\
hh:NaHC{O_3}(a\,mol);N{a_2}C{O_3}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 1\\
a - 1,5b = 0
\end{array} \right.\\
\Rightarrow a = 0,6;b = 0,4\\
{n_{NaOH}} = {n_{NaHC{O_3}}} + 2{n_{N{a_2}C{O_3}}} = 1,4mol\\
{V_{NaOH}} = \dfrac{{1,4}}{2} = 0,7l\\
\text{Để 2 dung dịch muối có nồng độ bằng nhau thì :}\\
\left\{ \begin{array}{l}
a + b = 1\\
a - b = 0
\end{array} \right.\\
\Rightarrow a = 0,5;b = 0,5\\
{n_{NaOH(1)}} = 0,5 + 0,5 \times 2 = 1,5mol\\
{V_{NaOH(1)}} = \dfrac{{1,5}}{2} = 0,75l\\
{V_{NaO{H_{ct}}}} = 0,75 - 0,7 = 0,05l = 50ml
\end{array}\)