\(\begin{array}{l}
a)\\
2{C_2}{H_2} + 5{O_2} \to 4C{O_2} + 2{H_2}O\\
b)\\
V{C_2}{H_2} = 20 \times 97\% = 19,4l\\
n{C_2}{H_2} = \dfrac{{19,4}}{{22,4}} = 0,866\,mol\\
= > n{O_2} = 0,866 \times \frac{5}{2} = 2,165\,mol\\
V{O_2} = 2,165 \times 22,4 = 48,496l\\
c)\\
nC{O_2} = 2n{C_2}{H_2} = 1,732mol\\
mC{O_2} = 1,732 \times 44 = 76,208g\\
n{H_2}O = n{C_2}{H_2} = 0,866\,mol\\
m{H_2}O = 0,866 \times 18 = 15,588g
\end{array}\)