1. Write the combustion reactions

For a hydrocarbon $C_xH_y$, complete combustion is: $C_xH_y + \left(x + \frac{y}{4}\right)O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O$

2. For propane, $C_3H_8$

Using the formula: $O_2 = 3 + \frac{8}{4} = 3 + 2 = 5$

So, $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$

Thus, 1 mole of propane requires 5 moles of $O_2$.

3. For butane, $C_4H_{10}$

Using the formula: $O_2 = 4 + \frac{10}{4} = 4 + 2.5 = 6.5$

So, $C_4H_{10} + \frac{13}{2}O_2 \rightarrow 4CO_2 + 5H_2O$

Thus, 1 mole of butane requires $\frac{13}{2}$ moles of $O_2$.

Therefore, 2 moles of butane require: $2 \times \frac{13}{2} = 13 \text{ moles of } O_2$

4. Total oxygen required

For 1 mole propane + 2 moles butane: $5 + 13 = 18$

Hence, the minimum number of moles of $O_2$ required is: $\boxed{18}$