Đáp án:
a)
\(\begin{array}{l}
a)\\
\text{Dung dịch NaOH dư}\\
{m_{NaO{H_d}}} = 1,6g\\
b)\\
{m_{N{a_2}C{O_3}}} = 14,84g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
C{O_2} + 2NaOH \to N{a_2}C{O_3} + {H_2}O\\
{n_{C{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{3,136}}{{22,4}} = 0,14mol\\
{n_{NaOH}} = \dfrac{m}{M} = \dfrac{{12,8}}{{40}} = 0,32mol\\
\dfrac{{0,14}}{1} < \dfrac{{0,32}}{2} \Rightarrow NaOH \text{ dư}\\
{n_{NaOHd}} = 0,32 - 0,14 \times 2 = 0,04mol\\
{m_{NaO{H_d}}} = n \times M = 0,04 \times 40 = 1,6g\\
b)\\
{n_{N{a_2}C{O_3}}} = {n_{C{O_2}}} = 0,14mol\\
{m_{N{a_2}C{O_3}}} = n \times M = 0,14 \times 106 = 14,84g
\end{array}\)