\(\begin{array}{l}
m_{CO_2}+m_{H_2O}=80(g)\\
Ma\,m_{CO_2}:m_{H_2O}=11:9\\
\to \frac{m_{CO_2}}{11}=\frac{m_{H_2O}}{9}=\frac{m_{CO_2}+m_{H_2O}}{11+9}=\frac{80}{20}=4\\
\to m_{CO_2}=44(g);m_{H_2O}=36(g)\\
\to n_{CO_2}=\frac{44}{44}=1(mol);n_{H_2O}=\frac{36}{18}=2(mol)\\
Bao\,toan\,KL:\\
a=m_{CO_2}+m_{H_2O}-m_{O_2}=80-64=16(g)
\end{array}\)