Answer : The correct option is, (C) 26.5 L
Explanation :
The combustion of ethylene is:
![C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(g)](https://tex.z-dn.net/?f=C_2H_4%28g%29%2B3O_2%28g%29%5Crightarrow%202CO_2%28g%29%2B2H_2O%28g%29)
First we have to calculate the number of moles of water vapor.
From the balanced chemical reaction, we conclude that:
As, 1 mole of
react to give 2 moles of ![H_2O](https://tex.z-dn.net/?f=H_2O)
So, 0.535 mole of
react to give
moles of ![H_2O](https://tex.z-dn.net/?f=H_2O)
Now we have to calculate the volume of water vapor.
Using ideal gas equation:
![PV=nRT\\\\V=\frac{nRT}{P}](https://tex.z-dn.net/?f=PV%3DnRT%5C%5C%5C%5CV%3D%5Cfrac%7BnRT%7D%7BP%7D)
where,
P = pressure of gas = 100 kPa
V = volume of gas = ?
T = temperature of gas = ![25^oC=273+25=298K](https://tex.z-dn.net/?f=25%5EoC%3D273%2B25%3D298K)
R = gas constant = 8.314 L.kPa/K.mol
n = number of moles of gas = 1.07 mol
Now put all the given values in the above formula, we get:
![V=\frac{nRT}{P}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7BnRT%7D%7BP%7D)
![V=\frac{1.07 mol\times 8.314 L.kPa/K.mol\times 298K}{100kPa}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B1.07%20mol%5Ctimes%208.314%20L.kPa%2FK.mol%5Ctimes%20298K%7D%7B100kPa%7D)
![V=26.5L](https://tex.z-dn.net/?f=V%3D26.5L)
Therefore, the volume of water vapor is, 26.5 L.