\(\begin{array}{l}
1)\\
nC{O_2} = \dfrac{m}{M} = \dfrac{{6,6}}{{44}} = 0,15\,mol\\
nC = nC{O_2} = 0,15\,mol\\
n{H_2}O = \dfrac{m}{M} = \frac{{2,7}}{{18}} = 0,15\,mol\\
nH = 2n{H_2}O = 0,3\,mol\\
mC + mH = 0,15 \times 12 + 0,3 = 2,1g\\
mO = 4,5 - 2,1 = 2,4g\\
nO = \dfrac{m}{M} = \frac{{2,4}}{{16}} = 0,15\,mol\\
nC:nH:nO = 0,15:0,3:0,15 = 1:2:1\\
 \Rightarrow CTDGN:C{H_2}O\\
MA = 60g/mol \Rightarrow 30n = 60 \Rightarrow n = 2\\
 \Rightarrow CTPT:{C_2}{H_4}{O_2}\\
2)\\
a)\\
{C_2}{H_4} + 3{O_2} \to 2C{O_2} + 2{H_2}O\\
n{C_2}{H_4} = \dfrac{V}{{22,4}} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
nC{O_2} = 2n{C_2}{H_4} = 0,4mol\\
VC{O_2} = n \times 22,4 = 0,4 \times 22,4 = 8,96l\\
b)\\
n{O_2} = 3n{C_2}{H_4} = 0,6\,mol\\
Vkk = 0,6 \times 5 \times 22,4 = 67,2l
\end{array}\)