At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% $$O_2$$ by volume, for complete combustion. After combustion, the gases occupy 345 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is: (Assume complete combustion of reactant)

We are given that 15 mL of an unknown gaseous hydrocarbon reacts completely with 375 mL of air at 300 K and 1 atm. The air contains 20 % $$O_2$$ by volume, so first we separate the oxygen and nitrogen present in this air sample.

20 % of 375 mL is oxygen, the rest is nitrogen:

$$V_{O_2\,(\,\text{present in air}\,)} = 0.20 \times 375\,\text{mL} = 75\,\text{mL}$$

$$V_{N_2\,(\,\text{present in air}\,)} = 375\,\text{mL} - 75\,\text{mL} = 300\,\text{mL}$$

Let the unknown hydrocarbon be $$C_xH_y.$$ The complete combustion is written as

$$C_xH_y + \left( x + \frac{y}{4} \right)O_2 \;\rightarrow\; xCO_2 + \frac{y}{2}H_2O$$

All volumes are measured at the same temperature and pressure, therefore volume ratios equal mole ratios. For 15 mL of the hydrocarbon, the stoichiometric oxygen required is

$$V_{O_2\,(\,\text{required}\,)}=15\,\text{mL}\times\left( x+\frac{y}{4} \right)=15\left(x+\frac{y}{4}\right)\,\text{mL}$$

The products that remain in the gaseous state after combustion (water is stated to condense) are $$CO_2,$$ any unused $$O_2,$$ and all the $$N_2.$$ Their total measured volume is 345 mL.

Initial total volume before ignition:

$$V_{\text{initial}} = 15\,\text{mL (hydrocarbon)} + 375\,\text{mL (air)} = 390\,\text{mL}$$

Final total volume after reaction (with liquid water removed):

$$V_{\text{final}} = 345\,\text{mL}$$

Hence the net contraction in volume is

$$\Delta V = V_{\text{initial}} - V_{\text{final}} = 390\,\text{mL} - 345\,\text{mL} = 45\,\text{mL}$$

This contraction can also be expressed in terms of the reacting gases. The nitrogen (300 mL) is inert, so its volume does not change. The hydrocarbon (15 mL) disappears, $$x$$ volumes of $$CO_2$$ for each volume of hydrocarbon appear, and $$O_2$$ is partly or completely consumed. We now count the final gaseous volumes one by one.

Final $$N_2$$ volume (unchanged): $$300\,\text{mL}.$$

Final $$CO_2$$ volume produced: each mole of hydrocarbon gives $$x$$ moles of $$CO_2,$$ so

$$V_{CO_2} = 15x\,\text{mL}$$

Final unused $$O_2$$ volume:

$$V_{O_2\,(\,\text{unused}\,)} = 75\,\text{mL} - 15\left(x+\frac{y}{4}\right)\,\text{mL}$$

Total final volume therefore is

$$V_{\text{final}} = V_{N_2} + V_{CO_2} + V_{O_2\,(\,\text{unused}\,)}$$

$$\phantom{V_{\text{final}}}=300 + 15x + \left[\,75 - 15\left(x+\frac{y}{4}\right)\right]$$

$$\phantom{V_{\text{final}}}=300 + 75 + 15x - 15x - \frac{15y}{4}$$

$$\phantom{V_{\text{final}}}=375 - \frac{15y}{4}$$

But we know that this must equal the experimentally observed 345 mL, so

$$375 - \frac{15y}{4} = 345$$

Subtracting 345 on both sides gives

$$30 = \frac{15y}{4}$$

Multiplying by 4 and dividing by 15 gives

$$y = 8$$

Thus the hydrocarbon contains eight hydrogen atoms.

Now we evaluate how much oxygen has actually been consumed when $$y=8.$$ The oxygen volume required becomes

$$V_{O_2\,(\,\text{required}\,)} = 15\left(x+\frac{8}{4}\right) = 15\left(x+2\right) = 15x + 30\,\text{mL}$$

The remaining oxygen (which cannot be negative) is

$$V_{O_2\,(\,\text{unused}\,)} = 75 - (15x + 30) = 45 - 15x\,\text{mL}$$

Because a negative unused volume is impossible, we need $$45 - 15x \ge 0,$$ hence

$$x \le 3$$

Possible integer values of $$x$$ are therefore 1, 2, or 3. Among the answer choices, the only formula with $$y = 8$$ and such an $$x$$ is $$C_3H_8.$$ (If $$x=3,$$ the unused oxygen volume becomes exactly zero, which is perfectly permissible.)

Consequently, the unknown hydrocarbon is propane, $$C_3H_8.$$

Hence, the correct answer is Option D.