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* 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=(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=(nRT)/(P)`

`V=(1.07 mol* 8.314 L.kPa/K.mol* 298K)/(100kPa)`

`V=26.5L`

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