\(\begin{array}{l}
1)\\
C{H_3}COONa + NaOH \to C{H_4} + N{a_2}C{O_3}\\
2C{H_4} \to {C_2}{H_2} + 3{H_2}\\
{C_2}{H_2} + {H_2} \to {C_2}{H_4}\\
{C_2}{H_4} + {H_2} \to {C_2}{H_6}\\
2{C_2}{H_2} \to {C_4}{H_4}\\
2)\\
nankan = n{H_2}O - nC{O_2} = 0,4 - 0,35 = 0,05\,mol\\
\% nankan = \dfrac{{0,05}}{{0,2}} \times 100\% = 25\%
\end{array}\)