Đáp án:

\(\begin{array}{l}
a)\\
{m_{{C_2}{H_2}}} = 6,5g\\
{m_{C{H_4}}} = 20g\\
b)\\
{m_{N{a_2}C{O_3}}} = 15,9g\\
{m_{NaHC{O_3}}} = 134,4g
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
{C_2}{H_2} + 2B{r_2} \to {C_2}{H_2}B{r_4}\\
{n_{B{r_2}}} = \dfrac{{80}}{{160}} = 0,5\,mol\\
{n_{{C_2}{H_2}}} = \dfrac{{0,5}}{2} = 0,25\,mol\\
{n_{hh}} = \dfrac{{33,6}}{{22,4}} = 1,5\,mol\\
{n_{C{H_4}}} = 1,5 - 0,25 = 1,25\,mol\\
{m_{{C_2}{H_2}}} = 0,25 \times 26 = 6,5g\\
{m_{C{H_4}}} = 1,25 \times 16 = 20g\\
b)\\
C{H_4} + 2{O_2} \xrightarrow{t^0} C{O_2} + 2{H_2}O\\
2{C_2}{H_2} + 5{O_2} \xrightarrow{t^0} 4C{O_2} + 2{H_2}O\\
{n_{C{O_2}}} = 1,25 + 0,25 \times 2 = 1,75\,mol\\
{n_{NaOH}} = 0,5 \times 3,8 = 1,9\,mol\\
T = \dfrac{{{n_{NaOH}}}}{{{n_{C{O_2}}}}} = \dfrac{{1,9}}{{1,75}} = 1,09\\
1 < T < 2 \Rightarrow \text{ Tạo cả 2 muối } \\
2NaOH + C{O_2} \to N{a_2}C{O_3} + {H_2}O(1)\\
N{a_2}C{O_3} + C{O_2} + {H_2}O \to 2NaHC{O_3}(2)\\
{n_{N{a_2}C{O_3}(1)}} = {n_{C{O_2}(1)}} = \dfrac{{1,9}}{2} = 0,95\,mol\\
{n_{N{a_2}C{O_3}(2)}} = {n_{C{O_2}(2)}} = 1,75 - 0,95 = 0,8\,mol\\
{n_{N{a_2}C{O_3}}} = 0,95 - 0,8 = 0,15\,mol\\
{n_{NaHC{O_3}}} = 0,8 \times 2 = 1,6\,mol\\
{m_{N{a_2}C{O_3}}} = 0,15 \times 106 = 15,9g\\
{m_{NaHC{O_3}}} = 1,6 \times 84 = 134,4g
\end{array}\)