Question

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 $330\, mL$. 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 :

• A. $C_3H_6$
• B. $C_3H_8$
• C. $C_4H_8$
• D. $C_4H_{10}$

Answer 1

The Correct Option is B

To find the formula of the hydrocarbon, we need to follow these steps:

1. Identify the volume of oxygen used for combustion. Given that the air consists of 20% oxygen by volume, calculate the volume of oxygen in 375 mL of air:
2. \text{Volume of } O_2 = 0.20 \times 375 \, \text{mL} = 75 \, \text{mL} \end{align*}.
3. Note that the entire hydrocarbon is burned using 75 mL of oxygen. Given 15 mL of hydrocarbon, we have the complete combustion reaction as follows:
4. C_xH_y + \left(x + \frac{y}{4} \right) O_2 \rightarrow x CO_2 + \frac{y}{2} H_2O.
5. According to Gay-Lussac's law of combining volumes (at constant temperature and pressure), the ratio of the volumes of gases is the same as that of their reacting mole numbers.
6. After combustion, you have the remaining 330 mL of gases. Assume the water formed is liquid, so it does not count in the gas volume.
7. Substitute the given volume to calculate the number of moles:
8. CO_2 + \frac{3}{2}N_2 = 330 \, \text{mL} - 75\, \text{mL of O}_2
9. Thus, there were 255 mL of gas (excluding oxygen consumed) after combustion which should correspond to the volume of CO₂:
   • The entire O₂ volume is used with no leftover combined with nitrogen from air composition, and you calculate rebound as:
   • Thus: 4x = 60\\ \Rightarrow 60 \, \text{mL} \, CO_2 + 27 \times 5 \, \text{mL} \, \text{for water residue}

Calculate number of carbons per this stoichiometry:

• The molar ratio further depends on stoichiometry: x+m=h\\ \Rightarrow 3+m=8