The minimum number of moles of $$\text{O}_2$$ required for complete combustion of 1 mole of propane and 2 moles of butane is...............

First, let us write the general balanced combustion equation for an alkane. For an alkane of the form $$\text{C}_n\text{H}_{2n+2}$$, complete combustion in oxygen is given by

$$\text{C}_n\text{H}_{2n+2} \;+\;\frac{3n+1}{2}\,\text{O}_2 \;\longrightarrow\; n\,\text{CO}_2 \;+\;(n+1)\,\text{H}_2\text{O}.$$

This relation comes from balancing carbon atoms first ($$n\;\text{CO}_2$$), hydrogen atoms next ($$(n+1)\;\text{H}_2\text{O}$$ provides $$2(n+1)$$ hydrogens), and finally oxygen atoms, giving the factor $$\dfrac{3n+1}{2}$$ in front of $$\text{O}_2$$.

Now we apply this result to each hydrocarbon present in the mixture.

1. Propane has $$n = 3$$, i.e. its formula is $$\text{C}_3\text{H}_8$$. Substituting $$n = 3$$ into the oxygen coefficient $$\dfrac{3n+1}{2}$$ gives

$$\dfrac{3(3)+1}{2} \;=\; \dfrac{9+1}{2} \;=\; \dfrac{10}{2} \;=\; 5.$$

So the balanced combustion reaction for propane is

$$\text{C}_3\text{H}_8 + 5\,\text{O}_2 \;\longrightarrow\; 3\,\text{CO}_2 + 4\,\text{H}_2\text{O}.$$

Hence, 1 mole of propane requires $$5$$ moles of $$\text{O}_2$$.

2. Butane has $$n = 4$$, i.e. its formula is $$\text{C}_4\text{H}_{10}$$. Substituting $$n = 4$$ into the same expression $$\dfrac{3n+1}{2}$$ gives

$$\dfrac{3(4)+1}{2} \;=\; \dfrac{12+1}{2} \;=\; \dfrac{13}{2} \;=\; 6.5.$$

Thus the balanced combustion reaction for butane is

$$\text{C}_4\text{H}_{10} + \dfrac{13}{2}\,\text{O}_2 \;\longrightarrow\; 4\,\text{CO}_2 + 5\,\text{H}_2\text{O}.$$

Therefore, 1 mole of butane needs $$6.5$$ moles of $$\text{O}_2$$. Since the question contains 2 moles of butane, the oxygen required for butane is

$$2 \times 6.5 = 13 \text{ moles of } \text{O}_2.$$

We now add the oxygen required for propane and for butane:

$$\text{Total } \text{O}_2 = 5 \;+\; 13 = 18 \text{ moles}.$$

So, the answer is $$18$$.