\(\begin{array}{l}
Bao\,toan\,C:\,n_C=n_{CO_2}=\frac{4,4}{44}=0,1(mol)\\
Bao\,toan\,H:\,n_H=2n_{H_2O}=2.\frac{0,9}{18}=0,1(mol)\\
\to m_C+m_H=0,1.12+0,1=1,3=m_A\\
\to A\,chua\,C,H\\
\to n_C:n_H=0,1:0,1=1:1\\
\to CTN:(CH)_n\\
Ma\,M_A=26(g/mol)\\
\to (12+1).n=26\\
\to n=2\\
\to A:C_2H_2
\end{array}\)