$\begin{align}
  & 2{{C}_{8}}{{H}_{18}}(l)+25{{O}_{2}}(g)\to 16C{{O}_{2}}(g)+18{{H}_{2}}O(l) \\
 & ?mol\,\,{{O}_{2}}=570mL\,\,{{C}_{8}}{{H}_{18}}\times \frac{0/8\,\,{{C}_{8}}{{H}_{18}}}{1mL\,\,{{C}_{8}}{{H}_{18}}}\times \frac{1\,mol\,\,{{C}_{8}}{{H}_{18}}}{114g\,\,{{C}_{8}}{{H}_{18}}}\times \frac{2\,5mol\,\,{{O}_{2}}}{2mol\,\,{{C}_{8}}{{H}_{18}}}=50mol\,\,{{O}_{2}} \\
 & {{C}_{2}}{{H}_{5}}OH(l)+3{{O}_{2}}(g)\to 2C{{O}_{2}}(g)+3{{H}_{2}}O(l) \\
 & ?\,mol\,\,{{O}_{2}}=430mL\,{{C}_{2}}{{H}_{5}}OH\times \frac{0/92g\,\,{{C}_{2}}{{H}_{5}}OH}{1\,mL\,\,{{C}_{2}}{{H}_{5}}OH}\,\times \frac{1\,mol\,\,{{C}_{2}}{{H}_{5}}OH}{46g\,\,{{C}_{2}}{{H}_{5}}OH}\times \frac{3mol\,\,{{O}_{2}}}{1mol\,\,{{C}_{2}}{{H}_{5}}OH}=25/8{{O}_{2}} \\
 & ?L\,Air=(50+25/8)mol\,\,{{O}_{2}}\times \frac{22/4L\,\,{{O}_{2}}}{1mol\,\,{{O}_{2}}}\times \frac{5L\,Air}{1L\,\,{{O}_{2}}}\simeq 8490L\,Air \\
\end{align}$