Partial Pressure of Gases Question! URGENT PLEASE!?

Apr 2013
15
0
Hi! I am new to this forum so bear with me please :)

I have this Inorganic A-Level Chemistry Question about Dalton's Law of Partial Pressure:

A mixture of gases containing 70cm3 of propane, C3H8 and 50cm3 of propene, C3H6, were exploded with 3.0dm3 of air containing 21% oxygen, at a pressure of 1x10^5 Pa. All volumes were measured at s.t.p.

(i) Write a balanced equations for the complete combustion of each gaseous hydrocarbon.

(ii) Calculate the partial pressure, in kPa, of propane, propene and oxygen in the original mixture of hydrocarbons.


I have figured out question (i), which is:

C3H8 + 5O2 --> 3CO2 + 4H2O
2C3H6 + 9O2 --> 6CO2 + 6H2O

Total: C3H8 + 2C3H6 + 14O2 --> 9CO2 + 10H2O

But, I can't do question (ii) :((((

Please help me :(

Thanks a lot! :)
 

Unknown008

CHF Hall of Fame
May 2010
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The gases were exploded at a pressure of 1x10^5 Pa.

You can use the ideal gas equation:

PV = nRT.

Taking R = 8.31 J/(molK)

For propane, we have
V = 70 cm^3, which is 70x10^-4 m^3 of gas,
P = 1x10^5 Pa
n = ?
R = 8.31
T = 273.15 K

You can find n of C3H8.

Do the same to get the number of moles of C3H6 and O2 and the remaining volume of air that is not oxygen.

Remember that partial pressure =


where n is the number of moles of the gas,
N is the total number of moles of all gases,
P is the total pressure of the gases together.

So, all the partial pressures should add up to 1x10^5 when you got each.

Could you try this and post what you get?
 
Last edited:
Apr 2013
15
0
Although, may I correct just one thing...

The temperature is not 298K but 273.15K because we are at s.t.p not r.t.p :)

But otherwise it was a perfect answer! :)
 

Unknown008

CHF Hall of Fame
May 2010
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Mauritius
Oh! I somehow saw rtp :eek: I'll fix it in my post above.

Glad I was able to help!