Question

I need help with question 5. Ignore my writing A 25.6 B 26.1 C 26.5 D 26.2

Answer 1

Answer : The correct option is, (C) 26.5 L

Explanation :

The combustion of ethylene is:

$C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(g)$

First we have to calculate the number of moles of water vapor.

From the balanced chemical reaction, we conclude that:

As, 1 mole of $C_2H_4$ react to give 2 moles of $H_2O$

So, 0.535 mole of $C_2H_4$ react to give $2\times 0.535=1.07mol$ moles of $H_2O$

Now we have to calculate the volume of water vapor.

Using ideal gas equation:

$PV=nRT\\\\V=\frac{nRT}{P}$

where,

P = pressure of gas = 100 kPa

V = volume of gas = ?

T = temperature of gas = $25^oC=273+25=298K$

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:

$V=\frac{nRT}{P}$

$V=\frac{1.07 mol\times 8.314 L.kPa/K.mol\times 298K}{100kPa}$

$V=26.5L$

Therefore, the volume of water vapor is, 26.5 L.