Answer : The correct option is, (C) 26.5 L
Explanation :
The combustion of ethylene is:
[tex]C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(g)[/tex]
First we have to calculate the number of moles of water vapor.
From the balanced chemical reaction, we conclude that:
As, 1 mole of [tex]C_2H_4[/tex] react to give 2 moles of [tex]H_2O[/tex]
So, 0.535 mole of [tex]C_2H_4[/tex] react to give [tex]2\times 0.535=1.07mol[/tex] moles of [tex]H_2O[/tex]
Now we have to calculate the volume of water vapor.
Using ideal gas equation:
[tex]PV=nRT\\\\V=\frac{nRT}{P}[/tex]
where,
P = pressure of gas = 100 kPa
V = volume of gas = ?
T = temperature of gas = [tex]25^oC=273+25=298K[/tex]
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:
[tex]V=\frac{nRT}{P}[/tex]
[tex]V=\frac{1.07 mol\times 8.314 L.kPa/K.mol\times 298K}{100kPa}[/tex]
[tex]V=26.5L[/tex]
Therefore, the volume of water vapor is, 26.5 L.
