Đáp án:

 C2H6

1,6 g

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
nC{O_2} = \dfrac{m}{M} = \dfrac{{17,6}}{{44}} = 0,4\,mol\\
nC = nC{O_2} = 0,4\,mol\\
n{H_2}O = \dfrac{m}{M} = \dfrac{{10,8}}{{18}} = 0,6\,mol\\
nH = 2n{H_2}O = 1,2\,mol\\
mC + mH = 0,4 \times 12 + 1,2 = 6g\\
 \Rightarrow A:C,H\\
b)\\
nC:nH = 0,4:1,2 = 1:3\\
 \Rightarrow CTDGN:C{H_3}\\
MA = 30g/mol \Rightarrow 15n = 30 \Rightarrow n = 2\\
 \Rightarrow CTPT:{C_2}{H_6}\\
c)\\
n{C_2}{H_6} = \dfrac{m}{M} = \dfrac{3}{{30}} = 0,1\,mol\\
n{O_2} = \dfrac{m}{M} = \dfrac{{12,8}}{{32}} = 0,4\,mol\\
2{C_2}{H_6} + 7{O_2} \xrightarrow{t^0} 4C{O_2} + 6{H_2}O\\
\dfrac{{0,1}}{2} < \dfrac{{0,4}}{7} \Rightarrow \text{ Oxi dư}\\
n{O_2} = 0,4 - 0,35 = 0,05\,mol\\
m{O_2} = n \times M = 0,05 \times 32 = 1,6g
\end{array}\)