$CH_4 + 2O_2 \to CO_2 + 2H_2O\\\hspace{0,2cm}a\hspace{1,2cm}2a\hspace{1,2cm}a\\C_4H_{10} + \dfrac{13}{2}O_2 \to 4CO_2 + 5H_2O\\\hspace{0,5cm}b\hspace{1,4cm}6,5b\hspace{1,2cm}4b$
$m_{CH_4,\,C_4H_{10}}=11,1\,(g)\\\to 16a+58b=11,1\,(1)\\m_{CO_2}=33\,(g)\\\to 44a+176b=33\,(2)\\\text{Từ (1), (2)}\to \begin{cases}16a+58b=11,1\\44a+176b=33\end{cases}\\\to\begin{cases}a=0,15\\b=0,15\end{cases}$
$\to n_{O_2}=2a+6,5b=1,275\,(mol)$
$\to m_{O_2}=1,275\times32=40,8\,(g)$